Key Concepts and Formulas

• Section Formula: If a point $D$ divides the line segment joining points $A$ and $B$ with position vectors $\vec{a}$ and $\vec{b}$ respectively, in the ratio $m:n$, then the position vector of $D$ is given by $\vec{d} = \frac{n\vec{a} + m\vec{b}}{m+n}$.
• Area of a Triangle using Position Vectors: If the position vectors of the vertices of a triangle are $\vec{a}$, $\vec{b}$, and $\vec{c}$, then the area of the triangle is given by $\frac{1}{2} |(\vec{b}-\vec{a}) \times (\vec{c}-\vec{a})|$. Alternatively, if we consider the origin to be one of the vertices, say $\vec{a} = \vec{0}$, then the area is $\frac{1}{2} |\vec{b} \times \vec{c}|$.
• Properties of Vector Cross Product: The cross product of two vectors is distributive and scalar multiplication is commutative. Also, $\vec{a} \times \vec{a} = \vec{0}$.

Step-by-Step Solution

Step 1: Define position vectors for the vertices of $\triangle PQR$. Let the position vectors of points $P$, $Q$, and $R$ be $\vec{p}$, $\vec{q}$, and $\vec{r}$ respectively. We can choose an arbitrary origin $O$.

Step 2: Express the position vectors of points A, B, and C using the section formula. We are given that points $A$, $B$, and $C$ are on the sides $QR$, $RP$, and $PQ$ respectively. The ratios are given as: $\frac{QA}{AR} = \frac{1}{2}$. This means $A$ divides $QR$ in the ratio $1:2$. Using the section formula, the position vector of $A$, $\vec{a}$, is: $\vec{a} = \frac{2\vec{q} + 1\vec{r}}{1+2} = \frac{2\vec{q} + \vec{r}}{3}$

$\frac{RB}{BP} = \frac{1}{2}$. This means $B$ divides $RP$ in the ratio $1:2$. Using the section formula, the position vector of $B$, $\vec{b}$, is: $\vec{b} = \frac{2\vec{r} + 1\vec{p}}{1+2} = \frac{2\vec{r} + \vec{p}}{3}$

$\frac{PC}{CQ} = \frac{1}{2}$. This means $C$ divides $PQ$ in the ratio $1:2$. Using the section formula, the position vector of $C$, $\vec{c}$, is: $\vec{c} = \frac{2\vec{p} + 1\vec{q}}{1+2} = \frac{2\vec{p} + \vec{q}}{3}$

Step 3: Calculate the area of $\triangle PQR$ in terms of position vectors. Let's consider the origin to be at point $P$, so $\vec{p} = \vec{0}$. Then the position vectors of $Q$ and $R$ are $\vec{q}$ and $\vec{r}$ respectively. The area of $\triangle PQR$ is given by: $Area(\triangle PQR) = \frac{1}{2} |\vec{q} \times \vec{r}|$

Step 4: Calculate the vectors representing the sides of $\triangle ABC$. To find the area of $\triangle ABC$, we need to calculate two adjacent side vectors, for example, $\vec{AB}$ and $\vec{AC}$. First, let's adjust the position vectors of $A$, $B$, and $C$ with respect to the origin at $P$ ($\vec{p} = \vec{0}$). $\vec{a} = \frac{2\vec{q} + \vec{r}}{3}$ $\vec{b} = \frac{2\vec{r} + \vec{0}}{3} = \frac{2\vec{r}}{3}$ $\vec{c} = \frac{2\vec{0} + \vec{q}}{3} = \frac{\vec{q}}{3}$

Now, calculate the side vectors: $\vec{AB} = \vec{b} - \vec{a} = \frac{2\vec{r}}{3} - \frac{2\vec{q} + \vec{r}}{3} = \frac{2\vec{r} - 2\vec{q} - \vec{r}}{3} = \frac{\vec{r} - 2\vec{q}}{3}$ $\vec{AC} = \vec{c} - \vec{a} = \frac{\vec{q}}{3} - \frac{2\vec{q} + \vec{r}}{3} = \frac{\vec{q} - 2\vec{q} - \vec{r}}{3} = \frac{-\vec{q} - \vec{r}}{3}$

Step 5: Calculate the area of $\triangle ABC$. The area of $\triangle ABC$ is given by: $Area(\triangle ABC) = \frac{1}{2} |\vec{AB} \times \vec{AC}|$ $Area(\triangle ABC) = \frac{1}{2} \left| \left(\frac{\vec{r} - 2\vec{q}}{3}\right) \times \left(\frac{-\vec{q} - \vec{r}}{3}\right) \right|$ $Area(\triangle ABC) = \frac{1}{2} \left| \frac{1}{9} (\vec{r} - 2\vec{q}) \times (-\vec{q} - \vec{r}) \right|$ $Area(\triangle ABC) = \frac{1}{18} | (\vec{r} - 2\vec{q}) \times (-\vec{q} - \vec{r}) |$ Expand the cross product: $(\vec{r} - 2\vec{q}) \times (-\vec{q} - \vec{r}) = \vec{r} \times (-\vec{q}) + \vec{r} \times (-\vec{r}) - 2\vec{q} \times (-\vec{q}) - 2\vec{q} \times (-\vec{r})$ Using properties of cross product ($\vec{a} \times \vec{b} = -\vec{b} \times \vec{a}$ and $\vec{a} \times \vec{a} = \vec{0}$): $= -\vec{r} \times \vec{q} - \vec{r} \times \vec{r} + 2\vec{q} \times \vec{q} + 2\vec{q} \times \vec{r}$ $= \vec{q} \times \vec{r} - \vec{0} + \vec{0} + 2\vec{q} \times \vec{r}$ $= 3(\vec{q} \times \vec{r})$ Substitute this back into the area formula: $Area(\triangle ABC) = \frac{1}{18} | 3(\vec{q} \times \vec{r}) | = \frac{3}{18} |\vec{q} \times \vec{r}| = \frac{1}{6} |\vec{q} \times \vec{r}|$

Step 6: Calculate the ratio of the areas. We need to find $\frac{Area(\triangle PQR)}{Area(\triangle ABC)}$. $\frac{Area(\triangle PQR)}{Area(\triangle ABC)} = \frac{\frac{1}{2} |\vec{q} \times \vec{r}|}{\frac{1}{6} |\vec{q} \times \vec{r}|}$ $\frac{Area(\triangle PQR)}{Area(\triangle ABC)} = \frac{1/2}{1/6} = \frac{1}{2} \times 6 = 3$

Let's recheck the calculations. Given: $\frac{QA}{AR} = \frac{1}{2}$, $\frac{RB}{BP} = \frac{1}{2}$, $\frac{PC}{CQ} = \frac{1}{2}$. This means $A$ divides $QR$ in ratio $1:2$, $B$ divides $RP$ in ratio $1:2$, $C$ divides $PQ$ in ratio $1:2$.

Let's use a slightly different approach for the area of $\triangle ABC$. Area($\triangle ABC$) = Area($\triangle PQR$) - Area($\triangle AQC$) - Area($\triangle CPA$) - Area($\triangle BRC$).

Let $Area(\triangle PQR) = \Delta$. We are given $\frac{QA}{AR} = \frac{1}{2}$. So, $AR = \frac{2}{3} QR$ and $QA = \frac{1}{3} QR$. We are given $\frac{RB}{BP} = \frac{1}{2}$. So, $BP = \frac{2}{3} RP$ and $RB = \frac{1}{3} RP$. We are given $\frac{PC}{CQ} = \frac{1}{2}$. So, $CQ = \frac{2}{3} PQ$ and $PC = \frac{1}{3} PQ$.

Consider the ratio of areas of triangles with the same height. $Area(\triangle PQR) = \Delta$. $Area(\triangle AQC)$: $A$ is on $QR$ and $C$ is on $PQ$. $\frac{Area(\triangle AQC)}{Area(\triangle PQR)} = \frac{AQ}{QR} \cdot \frac{QC}{QP}$. We are given $\frac{QA}{AR} = \frac{1}{2}$, so $\frac{AQ}{QR} = \frac{1}{3}$. We are given $\frac{PC}{CQ} = \frac{1}{2}$, so $\frac{CQ}{QP} = \frac{2}{3}$. Therefore, $Area(\triangle AQC) = \frac{1}{3} \cdot \frac{2}{3} Area(\triangle PQR) = \frac{2}{9} \Delta$.

$Area(\triangle CPA)$: $C$ is on $PQ$ and $A$ is on $QR$. $\frac{Area(\triangle CPA)}{Area(\triangle PQR)} = \frac{PC}{PQ} \cdot \frac{PA}{PR}$ (This is not directly applicable as $A$ is on $QR$). Let's use the given ratios directly. $\frac{PC}{CQ} = \frac{1}{2}$. This means $PC = \frac{1}{3} PQ$ and $CQ = \frac{2}{3} PQ$. $\frac{QA}{AR} = \frac{1}{2}$. This means $QA = \frac{1}{3} QR$ and $AR = \frac{2}{3} QR$. $\frac{RB}{BP} = \frac{1}{2}$. This means $RB = \frac{1}{3} RP$ and $BP = \frac{2}{3} RP$.

$Area(\triangle CPA)$: $C$ on $PQ$, $A$ on $QR$. The vertices are $C$, $P$, $A$. Consider $Area(\triangle CPA)$ relative to $Area(\triangle PQR)$. $Area(\triangle CPA) = \frac{1}{2} |\vec{p} \times \vec{c} + \vec{c} \times \vec{a} + \vec{a} \times \vec{p}|$. Let's use the subtraction method. $Area(\triangle ABC) = Area(\triangle PQR) - Area(\triangle AQC) - Area(\triangle CPA) - Area(\triangle BRC)$.

$Area(\triangle AQC)$: $A$ on $QR$, $C$ on $PQ$. $\frac{Area(\triangle AQC)}{Area(\triangle PQR)} = \frac{AQ}{QR} \cdot \frac{QC}{QP}$ is incorrect. It should be $\frac{AQ}{QR} \cdot \frac{PC}{PQ}$ if $C$ was on $PR$. The correct formula is: $Area(\triangle AQC) = \frac{1}{2} AQ \cdot QC \sin(\angle AQC)$. This is not helpful.

Let's use the property that if a point $D$ divides $AB$ in ratio $m:n$, then $Area(\triangle ADC)/Area(\triangle BDC) = m/n$. $Area(\triangle PQR) = \Delta$. $A$ is on $QR$ such that $\frac{QA}{AR} = \frac{1}{2}$. $Area(\triangle PQA) = \frac{1}{3} Area(\triangle PQR) = \frac{1}{3} \Delta$. $Area(\triangle PRA) = \frac{2}{3} Area(\triangle PQR) = \frac{2}{3} \Delta$.

$C$ is on $PQ$ such that $\frac{PC}{CQ} = \frac{1}{2}$. $Area(\triangle QCA) = \frac{2}{3} Area(\triangle PQA) = \frac{2}{3} \cdot \frac{1}{3} \Delta = \frac{2}{9} \Delta$. $Area(\triangle R CA) = \frac{2}{3} Area(\triangle PRA) = \frac{2}{3} \cdot \frac{2}{3} \Delta = \frac{4}{9} \Delta$. So, $Area(\triangle AQC) = \frac{2}{9} \Delta$.

$B$ is on $RP$ such that $\frac{RB}{BP} = \frac{1}{2}$. $Area(\triangle QBC) = \frac{1}{3} Area(\triangle QPC)$. $Area(\triangle QPC) = \frac{2}{3} Area(\triangle PQR) = \frac{2}{3} \Delta$. So, $Area(\triangle QBC) = \frac{1}{3} \cdot \frac{2}{3} \Delta = \frac{2}{9} \Delta$.

$Area(\triangle RBA) = \frac{1}{3} Area(\triangle RPA)$. $Area(\triangle RPA) = \frac{2}{3} Area(\triangle PQR) = \frac{2}{3} \Delta$. So, $Area(\triangle RBA) = \frac{1}{3} \cdot \frac{2}{3} \Delta = \frac{2}{9} \Delta$.

Let's re-evaluate the areas of the corner triangles. $Area(\triangle AQC)$. $A$ on $QR$, $C$ on $PQ$. The ratio of sides are: $\frac{QA}{QR} = \frac{1}{3}$ $\frac{QC}{QP} = \frac{2}{3}$ $Area(\triangle AQC) = \frac{AQ}{QR} \cdot \frac{QC}{QP} Area(\triangle PQR) = \frac{1}{3} \cdot \frac{2}{3} \Delta = \frac{2}{9} \Delta$. This is correct.

$Area(\triangle CPA)$. $C$ on $PQ$, $P$ is a vertex, $A$ on $QR$. The vertices are $C, P, A$. Consider $Area(\triangle CPA)$ relative to $Area(\triangle PQR)$. $\frac{Area(\triangle CPA)}{Area(\triangle PQR)} = \frac{PC}{PQ} \cdot \frac{PA}{PR}$ is not correct.

Let's use the vector cross product calculation again, as it is more systematic. Origin at P: $\vec{p} = \vec{0}$. $\vec{q}$, $\vec{r}$. $\vec{a} = \frac{2\vec{q} + \vec{r}}{3}$ $\vec{b} = \frac{2\vec{r}}{3}$ $\vec{c} = \frac{\vec{q}}{3}$

$\vec{AB} = \vec{b} - \vec{a} = \frac{2\vec{r}}{3} - \frac{2\vec{q} + \vec{r}}{3} = \frac{\vec{r} - 2\vec{q}}{3}$ $\vec{AC} = \vec{c} - \vec{a} = \frac{\vec{q}}{3} - \frac{2\vec{q} + \vec{r}}{3} = \frac{-\vec{q} - \vec{r}}{3}$

$Area(\triangle ABC) = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} \left| \left(\frac{\vec{r} - 2\vec{q}}{3}\right) \times \left(\frac{-\vec{q} - \vec{r}}{3}\right) \right|$ $= \frac{1}{18} |(\vec{r} - 2\vec{q}) \times (-\vec{q} - \vec{r})|$ $= \frac{1}{18} |-(\vec{r} \times \vec{q}) - (\vec{r} \times \vec{r}) + (2\vec{q} \times \vec{q}) + (2\vec{q} \times \vec{r})|$ $= \frac{1}{18} |\vec{q} \times \vec{r} + 0 - 0 + 2\vec{q} \times \vec{r}|$ $= \frac{1}{18} |3\vec{q} \times \vec{r}| = \frac{3}{18} |\vec{q} \times \vec{r}| = \frac{1}{6} |\vec{q} \times \vec{r}|$.

$Area(\triangle PQR) = \frac{1}{2} |\vec{q} \times \vec{r}|$. Ratio = $\frac{Area(\triangle PQR)}{Area(\triangle ABC)} = \frac{\frac{1}{2} |\vec{q} \times \vec{r}|}{\frac{1}{6} |\vec{q} \times \vec{r}|} = \frac{1/2}{1/6} = 3$.

Let's check the problem statement ratios again. QA/AR = 1/2. This means A divides QR in ratio 1:2. So $\vec{a} = \frac{2\vec{q} + 1\vec{r}}{3}$. This is correct. RB/BP = 1/2. This means B divides RP in ratio 1:2. So $\vec{b} = \frac{2\vec{r} + 1\vec{p}}{3}$. This is correct. PC/CQ = 1/2. This means C divides PQ in ratio 1:2. So $\vec{c} = \frac{2\vec{p} + 1\vec{q}}{3}$. This is correct.

The question asks for $\frac{Area(\triangle PQR)}{Area(\triangle ABC)}$. My calculation gives 3. The correct answer is A, which is 5/2. This indicates an error in my calculation or understanding of the problem.

Let's re-examine the question and ratios. $QA/AR = 1/2$. This means $AR = 2 QA$. So $QR = QA + AR = QA + 2QA = 3QA$. Thus $QA/QR = 1/3$ and $AR/QR = 2/3$. $RB/BP = 1/2$. This means $BP = 2 RB$. So $RP = RB + BP = RB + 2RB = 3RB$. Thus $RB/RP = 1/3$ and $BP/RP = 2/3$. $PC/CQ = 1/2$. This means $CQ = 2 PC$. So $PQ = PC + CQ = PC + 2PC = 3PC$. Thus $PC/PQ = 1/3$ and $CQ/PQ = 2/3$.

Okay, the ratios used in the vector section formula were correct. $\vec{a} = \frac{2\vec{q} + 1\vec{r}}{1+2} = \frac{2\vec{q} + \vec{r}}{3}$. This implies A divides QR in ratio 1:2, meaning QA/AR = 1/2. This is correct.

Let's use the area subtraction method again. Let $Area(\triangle PQR) = \Delta$. $Area(\triangle AQC) = \frac{AQ}{QR} \cdot \frac{QC}{QP} Area(\triangle PQR)$ is incorrect. This formula applies when the points are on adjacent sides originating from a common vertex.

Consider triangle PQR. A is on QR such that QA/AR = 1/2. So AQ/QR = 1/3. C is on PQ such that PC/CQ = 1/2. So CQ/PQ = 2/3. $Area(\triangle AQC) = \frac{1}{2} AQ \cdot QC \sin(\angle AQC)$. Not helpful.

Let's use the formula for the area of the inner triangle in terms of the outer triangle and the ratios. If points A, B, C divide sides QR, RP, PQ in ratios $m:n$, $p:q$, $r:s$ respectively, i.e., $QA/AR = m/n$, $RB/BP = p/q$, $PC/CQ = r/s$. Then $Area(\triangle ABC) = Area(\triangle PQR) \left( 1 - \frac{m}{m+n} \frac{s}{r+s} - \frac{p}{p+q} \frac{n}{m+n} - \frac{r}{r+s} \frac{q}{p+q} \right)$. This formula is for a different division.

Let's use the ratios as given: $QA/AR = 1/2 \implies AR = 2 QA$. $QR = 3 QA$. $AQ/QR = 1/3$. $RB/BP = 1/2 \implies BP = 2 RB$. $RP = 3 RB$. $RB/RP = 1/3$. $PC/CQ = 1/2 \implies CQ = 2 PC$. $PQ = 3 PC$. $PC/PQ = 1/3$.

Area of $\triangle ABC$ = Area of $\triangle PQR$ - Area of $\triangle AQC$ - Area of $\triangle CPA$ - Area of $\triangle BRC$. Let $Area(\triangle PQR) = \Delta$. $Area(\triangle AQC)$: A is on QR, C is on PQ. $\frac{Area(\triangle AQC)}{Area(\triangle PQR)} = \frac{AQ}{QR} \cdot \frac{CQ}{PQ}$ is incorrect.

The correct formula for $Area(\triangle AQC)$ when $A$ is on $QR$ and $C$ is on $PQ$: $Area(\triangle AQC) = \frac{AQ}{QR} \cdot \frac{PC}{PQ} Area(\triangle PQR)$ is incorrect.

Let's go back to the vector calculation and check for any fundamental errors. Origin at $P$: $\vec{p} = \vec{0}$. $\vec{a} = \frac{2\vec{q} + \vec{r}}{3}$ (A divides QR in 1:2) $\vec{b} = \frac{2\vec{r}}{3}$ (B divides RP in 1:2, since $\vec{p}=\vec{0}$) $\vec{c} = \frac{\vec{q}}{3}$ (C divides PQ in 1:2, since $\vec{p}=\vec{0}$)

$\vec{AB} = \vec{b} - \vec{a} = \frac{2\vec{r}}{3} - \frac{2\vec{q} + \vec{r}}{3} = \frac{\vec{r} - 2\vec{q}}{3}$. $\vec{AC} = \vec{c} - \vec{a} = \frac{\vec{q}}{3} - \frac{2\vec{q} + \vec{r}}{3} = \frac{-\vec{q} - \vec{r}}{3}$.

$Area(\triangle ABC) = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} \left| \frac{\vec{r} - 2\vec{q}}{3} \times \frac{-\vec{q} - \vec{r}}{3} \right|$ $= \frac{1}{18} |(\vec{r} - 2\vec{q}) \times (-\vec{q} - \vec{r})|$ $= \frac{1}{18} |-\vec{r}\times\vec{q} - \vec{r}\times\vec{r} + 2\vec{q}\times\vec{q} + 2\vec{q}\times\vec{r}|$ $= \frac{1}{18} |\vec{q}\times\vec{r} - 0 + 0 + 2\vec{q}\times\vec{r}| = \frac{1}{18} |3\vec{q}\times\vec{r}| = \frac{1}{6} |\vec{q}\times\vec{r}|$.

$Area(\triangle PQR) = \frac{1}{2} |\vec{q} \times \vec{r}|$. Ratio is 3.

Let's consider the case where the ratios are $1:k$. If $QA/AR = 1/k$, $RB/BP = 1/k$, $PC/CQ = 1/k$. Then $\vec{a} = \frac{k\vec{q} + \vec{r}}{k+1}$, $\vec{b} = \frac{k\vec{r} + \vec{p}}{k+1}$, $\vec{c} = \frac{k\vec{p} + \vec{q}}{k+1}$. With origin at P: $\vec{p} = \vec{0}$. $\vec{a} = \frac{k\vec{q} + \vec{r}}{k+1}$ $\vec{b} = \frac{k\vec{r}}{k+1}$ $\vec{c} = \frac{\vec{q}}{k+1}$

$\vec{AB} = \vec{b} - \vec{a} = \frac{k\vec{r}}{k+1} - \frac{k\vec{q} + \vec{r}}{k+1} = \frac{(k-1)\vec{r} - k\vec{q}}{k+1}$. $\vec{AC} = \vec{c} - \vec{a} = \frac{\vec{q}}{k+1} - \frac{k\vec{q} + \vec{r}}{k+1} = \frac{(1-k)\vec{q} - \vec{r}}{k+1}$.

$Area(\triangle ABC) = \frac{1}{2} |\vec{AB} \times \vec{AC}|$ $= \frac{1}{2} \left| \frac{(k-1)\vec{r} - k\vec{q}}{k+1} \times \frac{(1-k)\vec{q} - \vec{r}}{k+1} \right|$ $= \frac{1}{2(k+1)^2} | ((k-1)\vec{r} - k\vec{q}) \times ((1-k)\vec{q} - \vec{r}) |$ $= \frac{1}{2(k+1)^2} | -(k-1)^2 \vec{r}\times\vec{q} - (k-1)\vec{r}\times\vec{r} + k(1-k)\vec{q}\times\vec{q} + k\vec{q}\times\vec{r} |$ $= \frac{1}{2(k+1)^2} | (k-1)^2 \vec{q}\times\vec{r} + 0 + 0 + k\vec{q}\times\vec{r} |$ $= \frac{1}{2(k+1)^2} | (k^2 - 2k + 1 + k) \vec{q}\times\vec{r} |$ $= \frac{1}{2(k+1)^2} | (k^2 - k + 1) \vec{q}\times\vec{r} |$.

$Area(\triangle PQR) = \frac{1}{2} |\vec{q}\times\vec{r}|$. Ratio = $\frac{Area(\triangle PQR)}{Area(\triangle ABC)} = \frac{\frac{1}{2} |\vec{q}\times\vec{r}|}{\frac{1}{2(k+1)^2} (k^2 - k + 1) |\vec{q}\times\vec{r}|} = \frac{(k+1)^2}{k^2 - k + 1}$.

In our problem, $k=2$. Ratio = $\frac{(2+1)^2}{2^2 - 2 + 1} = \frac{3^2}{4 - 2 + 1} = \frac{9}{3} = 3$. This still gives 3.

Let's check the problem statement again. ${{QA} \over {AR}} = {{RB} \over {BP}} = {{PC} \over {CQ}} = {1 \over 2}$ This is $QA/AR = 1/2$. My interpretation was correct.

Let's consider the areas using the formula: Area of inner triangle $ABC$ is given by $Area(ABC) = Area(PQR) \left( 1 - \frac{AR}{QR} \frac{BP}{RP} \frac{CQ}{PQ} - \frac{QA}{QR} \frac{RB}{RP} \frac{PC}{PQ} \right)$ This formula is for specific divisions.

Let's use the subtraction method with correct area calculations. $Area(\triangle PQR) = \Delta$. $A$ on $QR$ with $QA/AR = 1/2 \implies AQ/QR = 1/3$. $B$ on $RP$ with $RB/BP = 1/2 \implies RB/RP = 1/3$. $C$ on $PQ$ with $PC/CQ = 1/2 \implies PC/PQ = 1/3$.

$Area(\triangle RBP) = \frac{RB}{RP} \cdot \frac{BP}{RP} Area(\triangle PQR)$ is incorrect.

$Area(\triangle RBP) = \frac{RB}{RP} Area(\triangle RPQ) = \frac{1}{3} Area(\triangle PQR) = \frac{1}{3} \Delta$. This is incorrect. $Area(\triangle RBP) = \frac{1}{2} RB \cdot BP \sin(\angle R)$. This is not helpful.

Consider the areas of the three corner triangles: $\triangle AQC$, $\triangle CPA$, $\triangle BRC$. $Area(\triangle AQC) = \frac{AQ}{QR} \cdot \frac{PC}{PQ} Area(\triangle PQR)$ is incorrect.

The ratio of areas of triangles with a common angle is the product of the ratios of the sides containing the angle. $Area(\triangle CQA) = \frac{CQ}{PQ} \cdot \frac{QA}{QR} Area(\triangle PQR)$. We have $CQ/PQ = 2/3$ and $QA/QR = 1/3$. So $Area(\triangle CQA) = \frac{2}{3} \cdot \frac{1}{3} \Delta = \frac{2}{9} \Delta$.

$Area(\triangle APB) = \frac{AP}{RP} \cdot \frac{PB}{PQ} Area(\triangle PQR)$ is incorrect.

Let's use the ratios given: QA/AR = 1/2 => AR = 2QA. QR = 3QA. AQ/QR = 1/3. AR/QR = 2/3. RB/BP = 1/2 => BP = 2RB. RP = 3RB. RB/RP = 1/3. BP/RP = 2/3. PC/CQ = 1/2 => CQ = 2PC. PQ = 3PC. PC/PQ = 1/3. CQ/PQ = 2/3.

Area($\triangle PQR$) = $\Delta$. Area($\triangle AQC$) = $\frac{AQ}{QR} \cdot \frac{CQ}{PQ} Area(\triangle PQR)$ is incorrect.

Let's consider triangle PQR. Point A is on QR such that $QA = \frac{1}{3} QR$. Point B is on RP such that $RB = \frac{1}{3} RP$. Point C is on PQ such that $PC = \frac{1}{3} PQ$.

Area($\triangle CPA$) = $\frac{PC}{PQ} Area(\triangle PQA)$ is incorrect.

Let's use the formula for the area of the inner triangle formed by joining points on the sides. If points $D, E, F$ divide $BC, CA, AB$ in ratios $l:1, m:1, n:1$, then $Area(DEF)/Area(ABC) = \frac{(lmn-1)^2}{(lm+l+1)(mn+m+1)(nl+n+1)}$. This is not the current problem.

Let's use the property of areas when a line segment divides a triangle. Consider $\triangle PQR$. $C$ is on $PQ$ such that $PC/CQ = 1/2$. So $PC/PQ = 1/3$. $Area(\triangle PRC) = \frac{PC}{PQ} Area(\triangle PQR) = \frac{1}{3} \Delta$. $Area(\triangle QRC) = \frac{QC}{PQ} Area(\triangle PQR) = \frac{2}{3} \Delta$.

Now consider $\triangle QRC$. $A$ is on $QR$ such that $QA/AR = 1/2$. So $AR/QR = 2/3$. $Area(\triangle ARC) = \frac{AR}{QR} Area(\triangle QRC) = \frac{2}{3} \cdot \frac{2}{3} \Delta = \frac{4}{9} \Delta$.

Consider $\triangle PQC$. $A$ is on $QR$. This is not helpful.

Let's use the formula directly: If $QA/AR = x$, $RB/BP = y$, $PC/CQ = z$. Then $\frac{Area(\triangle ABC)}{Area(\triangle PQR)} = 1 - \frac{x}{(x+1)} \frac{1}{(z+1)} - \frac{y}{(y+1)} \frac{1}{(x+1)} - \frac{z}{(z+1)} \frac{1}{(y+1)}$. This is not correct.

The formula for the ratio of areas when points divide the sides is: Let $QA/AR = m/n$, $RB/BP = p/q$, $PC/CQ = r/s$. $Area(ABC)/Area(PQR) = 1 - \frac{m}{m+n}\frac{s}{r+s} - \frac{p}{p+q}\frac{n}{m+n} - \frac{r}{r+s}\frac{q}{p+q}$. Here $m=1, n=2$, so $QA/AR = 1/2$. $p=1, q=2$, so $RB/BP = 1/2$. $r=1, s=2$, so $PC/CQ = 1/2$.

$Area(ABC)/Area(PQR) = 1 - \frac{1}{1+2}\frac{2}{1+2} - \frac{1}{1+2}\frac{2}{1+2} - \frac{1}{1+2}\frac{2}{1+2}$ $Area(ABC)/Area(PQR) = 1 - \frac{1}{3}\frac{2}{3} - \frac{1}{3}\frac{2}{3} - \frac{1}{3}\frac{2}{3}$ $Area(ABC)/Area(PQR) = 1 - \frac{2}{9} - \frac{2}{9} - \frac{2}{9} = 1 - \frac{6}{9} = 1 - \frac{2}{3} = \frac{1}{3}$.

So $Area(\triangle PQR)/Area(\triangle ABC) = 3$. This still gives 3.

Let's re-read the question carefully. ${{QA} \over {AR}} = {{RB} \over {BP}} = {{PC} \over {CQ}} = {1 \over 2}$ This means A divides QR in ratio 1:2. B divides RP in ratio 1:2. C divides PQ in ratio 1:2.

Consider the case where the ratio is $1:2$ for all sides. Let the vertices of the outer triangle be $P=(0,0)$, $Q=(3,0)$, $R=(0,3)$. Area = $\frac{1}{2} \times 3 \times 3 = \frac{9}{2}$. A divides QR in ratio 1:2. $Q=(3,0), R=(0,3)$. $A = \frac{2(3,0) + 1(0,3)}{1+2} = \frac{(6,0) + (0,3)}{3} = \frac{(6,3)}{3} = (2,1)$. B divides RP in ratio 1:2. $R=(0,3), P=(0,0)$. $B = \frac{2(0,3) + 1(0,0)}{1+2} = \frac{(0,6) + (0,0)}{3} = \frac{(0,6)}{3} = (0,2)$. C divides PQ in ratio 1:2. $P=(0,0), Q=(3,0)$. $C = \frac{2(0,0) + 1(3,0)}{1+2} = \frac{(0,0) + (3,0)}{3} = \frac{(3,0)}{3} = (1,0)$.

Now, find the area of $\triangle ABC$. $A=(2,1), B=(0,2), C=(1,0)$. Area($\triangle ABC$) = $\frac{1}{2} |x_A(y_B-y_C) + x_B(y_C-y_A) + x_C(y_A-y_B)|$ $= \frac{1}{2} |2(2-0) + 0(0-1) + 1(1-2)|$ $= \frac{1}{2} |2(2) + 0 + 1(-1)|$ $= \frac{1}{2} |4 - 1| = \frac{3}{2}$.

Ratio = $\frac{Area(\triangle PQR)}{Area(\triangle ABC)} = \frac{9/2}{3/2} = \frac{9}{3} = 3$. This still gives 3.

Let's check the problem statement and the options again. Options are 5/2, 4, 2, 3. The correct answer is A, which is 5/2.

There must be a misunderstanding of the ratio division. $QA/AR = 1/2$. This means A is closer to Q. If Q is at origin, R is at vector $\vec{r}$. A is at $\frac{1}{3}\vec{r}$. The formulation $\vec{a} = \frac{n\vec{q} + m\vec{r}}{m+n}$ is for division in ratio $m:n$. So $QA/AR = m/n$. If $QA/AR = 1/2$, then $m=1, n=2$. $\vec{a} = \frac{2\vec{q} + 1\vec{r}}{1+2} = \frac{2\vec{q} + \vec{r}}{3}$. This is correct.

Let's re-examine the formula for the area of the inner triangle. If points A, B, C divide sides QR, RP, PQ such that $QA/AR = r_1$, $RB/BP = r_2$, $PC/CQ = r_3$. Then $Area(ABC)/Area(PQR) = \frac{(r_1 r_2 r_3 - 1)^2}{(r_1 r_2 + r_1 + 1)(r_2 r_3 + r_2 + 1)(r_3 r_1 + r_3 + 1)}$. This is for a different division.

Consider the case where the points divide the sides in the ratio $1:k$. $QA/AR = 1/k$. $Area(ABC)/Area(PQR) = \frac{(k+1)^2}{k^2-k+1}$. Here $k=2$. Ratio is 3.

Let's consider if the ratios are interpreted differently. Maybe the question means $AR/QR = 1/2$, $BP/RP = 1/2$, $CQ/PQ = 1/2$. If $AR/QR = 1/2$, then $A$ is the midpoint of QR. If $BP/RP = 1/2$, then $B$ is the midpoint of RP. If $CQ/PQ = 1/2$, then $C$ is the midpoint of PQ. In this case, Area(ABC)/Area(PQR) = 1/4. The ratio would be 4. This is option B.

Let's check the wording carefully: $QA/AR = 1/2$. This means the segment QR is divided into 3 parts, where QA is 1 part and AR is 2 parts. So $QA = \frac{1}{3} QR$ and $AR = \frac{2}{3} QR$.

Let's try the area subtraction method again, ensuring the ratios are used correctly. $Area(\triangle PQR) = \Delta$. $A$ on $QR$ such that $QA/AR = 1/2$. So $AQ/QR = 1/3$. $B$ on $RP$ such that $RB/BP = 1/2$. So $RB/RP = 1/3$. $C$ on $PQ$ such that $PC/CQ = 1/2$. So $PC/PQ = 1/3$.

$Area(\triangle CQA) = \frac{CQ}{PQ} \cdot \frac{QA}{QR} Area(\triangle PQR)$. $CQ/PQ = 2/3$. $QA/QR = 1/3$. $Area(\triangle CQA) = \frac{2}{3} \cdot \frac{1}{3} \Delta = \frac{2}{9} \Delta$.

$Area(\triangle APB) = \frac{AP}{RP} \cdot \frac{PB}{PQ} Area(\triangle PQR)$. $AP/RP$ is not directly given.

Let's use the ratios directly as given. QA/AR = 1/2. RB/BP = 1/2. PC/CQ = 1/2.

Consider the vector approach again, but with a different origin. Let the origin be $O$. $\vec{a} = \frac{2\vec{q} + \vec{r}}{3}$ $\vec{b} = \frac{2\vec{r} + \vec{p}}{3}$ $\vec{c} = \frac{2\vec{p} + \vec{q}}{3}$

$Area(\triangle ABC) = \frac{1}{2} |(\vec{b}-\vec{a}) \times (\vec{c}-\vec{a})|$ $\vec{b}-\vec{a} = \frac{2\vec{r} + \vec{p}}{3} - \frac{2\vec{q} + \vec{r}}{3} = \frac{\vec{p} - 2\vec{q} + \vec{r}}{3}$ $\vec{c}-\vec{a} = \frac{2\vec{p} + \vec{q}}{3} - \frac{2\vec{q} + \vec{r}}{3} = \frac{2\vec{p} - \vec{q} - \vec{r}}{3}$

$(\vec{b}-\vec{a}) \times (\vec{c}-\vec{a}) = \frac{1}{9} (\vec{p} - 2\vec{q} + \vec{r}) \times (2\vec{p} - \vec{q} - \vec{r})$ $= \frac{1}{9} [ \vec{p}\times(2\vec{p}) + \vec{p}\times(-\vec{q}) + \vec{p}\times(-\vec{r}) -2\vec{q}\times(2\vec{p}) -2\vec{q}\times(-\vec{q}) -2\vec{q}\times(-\vec{r}) + \vec{r}\times(2\vec{p}) + \vec{r}\times(-\vec{q}) + \vec{r}\times(-\vec{r}) ]$ $= \frac{1}{9} [ \vec{0} - \vec{p}\times\vec{q} - \vec{p}\times\vec{r} + 4\vec{p}\times\vec{q} + \vec{0} + 2\vec{q}\times\vec{r} + 2\vec{p}\times\vec{r} - \vec{r}\times\vec{q} + \vec{0} ]$ $= \frac{1}{9} [ 3\vec{p}\times\vec{q} + \vec{p}\times\vec{r} + 2\vec{q}\times\vec{r} + \vec{q}\times\vec{r} ]$ $= \frac{1}{9} [ 3\vec{p}\times\vec{q} + \vec{p}\times\vec{r} + 3\vec{q}\times\vec{r} ]$ $= \frac{1}{9} [ 3(\vec{p}\times\vec{q} + \vec{q}\times\vec{r}) + \vec{p}\times\vec{r} ]$ $= \frac{1}{9} [ 3(\vec{p}\times\vec{q} + \vec{q}\times\vec{r} + \vec{r}\times\vec{p}) ]$ (since $\vec{p}\times\vec{r} = -\vec{r}\times\vec{p}$) $= \frac{1}{3} (\vec{p}\times\vec{q} + \vec{q}\times\vec{r} + \vec{r}\times\vec{p})$.

$Area(\triangle PQR) = \frac{1}{2} |(\vec{q}-\vec{p}) \times (\vec{r}-\vec{p})|$ $= \frac{1}{2} |\vec{q}\times\vec{r} - \vec{q}\times\vec{p} - \vec{p}\times\vec{r} + \vec{p}\times\vec{p}|$ $= \frac{1}{2} |\vec{q}\times\vec{r} + \vec{p}\times\vec{q} + \vec{r}\times\vec{p}|$ $= \frac{1}{2} |\vec{p}\times\vec{q} + \vec{q}\times\vec{r} + \vec{r}\times\vec{p}|$.

So, $Area(\triangle ABC) = \frac{1}{3} Area(\triangle PQR)$. The ratio is 3.

Let's consider a different interpretation of the ratios. What if the question meant: $QA:AR = 1:2$ $AR:RP = 1:2$ (This is not possible as A is on QR)

Let's check the formula for the area ratio again. If $QA/AR = x$, $RB/BP = y$, $PC/CQ = z$. $Area(ABC)/Area(PQR) = \frac{(xyz-1)^2}{(xy+x+1)(yz+y+1)(zx+z+1)}$. This is for a specific type of division.

Let's use the result from a known theorem. If points A, B, C divide the sides QR, RP, PQ of triangle PQR in the ratio $1:k$ (i.e., $QA/AR = 1/k$, $RB/BP = 1/k$, $PC/CQ = 1/k$), then the area of triangle ABC is given by: $Area(\triangle ABC) = \frac{k^2-k+1}{(k+1)^2} Area(\triangle PQR)$. In our case, $QA/AR = 1/2$, so $k=2$. $Area(\triangle ABC) = \frac{2^2-2+1}{(2+1)^2} Area(\triangle PQR) = \frac{4-2+1}{3^2} Area(\triangle PQR) = \frac{3}{9} Area(\triangle PQR) = \frac{1}{3} Area(\triangle PQR)$. The ratio $Area(\triangle PQR)/Area(\triangle ABC) = 3$.

There might be an error in the provided correct answer or the question itself. However, let's consider another standard formula. If points A, B, C divide the sides QR, RP, PQ in the ratio $m:n$ (i.e., $QA/AR = m/n$, $RB/BP = m/n$, $PC/CQ = m/n$), then $Area(ABC)/Area(PQR) = \frac{(m^2 - mn + n^2)}{(m+n)^2}$. In our case, $QA/AR = 1/2$, so $m=1, n=2$. $Area(ABC)/Area(PQR) = \frac{1^2 - 1(2) + 2^2}{(1+2)^2} = \frac{1 - 2 + 4}{3^2} = \frac{3}{9} = \frac{1}{3}$. This again gives a ratio of 3.

Let's check if the ratios are applied in reverse. If $AR/QA = 1/2$, $BP/RB = 1/2$, $CQ/PC = 1/2$. This means $QA=2AR$, $RB=2BP$, $PC=2CQ$. $QR = QA+AR = 2AR+AR = 3AR$. So $AR/QR = 1/3$. $RP = RB+BP = 2BP+BP = 3BP$. So $BP/RP = 1/3$. $PQ = PC+CQ = 2CQ+CQ = 3CQ$. So $CQ/PQ = 1/3$. This means A, B, C are points such that $AR/QR=1/3$, $BP/RP=1/3$, $CQ/PQ=1/3$. This is the same as $QA/QR=2/3$, $RB/RP=2/3$, $PC/PQ=2/3$.

Let's consider the formula by Ceva's and Menelaus' theorem.

Let's re-check the provided correct answer: A, which is 5/2. This implies that the ratio $Area(\triangle PQR) / Area(\triangle ABC) = 5/2$. So $Area(\triangle ABC) = \frac{2}{5} Area(\triangle PQR)$.

Consider the formula: $Area(ABC)/Area(PQR) = 1 - \frac{AR}{QR}\frac{BP}{RP}\frac{CQ}{PQ} - \frac{QA}{QR}\frac{RB}{RP}\frac{PC}{PQ}$. This is incorrect.

Let's use the formula: Area($\triangle ABC$) = Area($\triangle PQR$) - Area($\triangle AQC$) - Area($\triangle CPA$) - Area($\triangle BRC$). $QA/AR = 1/2 \implies QA/QR = 1/3, AR/QR = 2/3$. $RB/BP = 1/2 \implies RB/RP = 1/3, BP/RP = 2/3$. $PC/CQ = 1/2 \implies PC/PQ = 1/3, CQ/PQ = 2/3$.

$Area(\triangle AQC) = \frac{AQ}{QR} \cdot \frac{CQ}{PQ} Area(\triangle PQR)$ is wrong. The correct formula for area of corner triangles is: $Area(\triangle CQA) = \frac{CQ}{PQ} \cdot \frac{QA}{QR} Area(\triangle PQR)$ if C is on PQ and A is on QR. $Area(\triangle CQA) = \frac{2}{3} \cdot \frac{1}{3} \Delta = \frac{2}{9} \Delta$.

$Area(\triangle APB) = \frac{AP}{RP} \cdot \frac{PB}{PQ} Area(\triangle PQR)$. Not directly applicable.

Let's use the vector method again, very carefully. Origin at P: $\vec{p} = \vec{0}$. $\vec{a} = \frac{2\vec{q} + \vec{r}}{3}$ $\vec{b} = \frac{2\vec{r}}{3}$ $\vec{c} = \frac{\vec{q}}{3}$

$\vec{AB} = \vec{b} - \vec{a} = \frac{2\vec{r}}{3} - \frac{2\vec{q} + \vec{r}}{3} = \frac{\vec{r} - 2\vec{q}}{3}$ $\vec{AC} = \vec{c} - \vec{a} = \frac{\vec{q}}{3} - \frac{2\vec{q} + \vec{r}}{3} = \frac{-\vec{q} - \vec{r}}{3}$

Area($\triangle ABC$) = $\frac{1}{2} |\vec{AB} \times \vec{AC}|$ $= \frac{1}{2} \left| \frac{\vec{r} - 2\vec{q}}{3} \times \frac{-\vec{q} - \vec{r}}{3} \right|$ $= \frac{1}{18} | (\vec{r} - 2\vec{q}) \times (-\vec{q} - \vec{r}) |$ $= \frac{1}{18} | -\vec{r}\times\vec{q} - \vec{r}\times\vec{r} + 2\vec{q}\times\vec{q} + 2\vec{q}\times\vec{r} |$ $= \frac{1}{18} | \vec{q}\times\vec{r} - 0 + 0 + 2\vec{q}\times\vec{r} | = \frac{1}{18} |3\vec{q}\times\vec{r}| = \frac{1}{6} |\vec{q}\times\vec{r}|$.

Area($\triangle PQR$) = $\frac{1}{2} |\vec{q} \times \vec{r}|$. Ratio = $\frac{Area(\triangle PQR)}{Area(\triangle ABC)} = \frac{1/2}{1/6} = 3$.

There seems to be a consistent result of 3, which is option D. However, the provided answer is A (5/2). Let's consider the possibility that the ratios are applied in a different order or meaning.

If $QA/AR = 2/1$, $RB/BP = 2/1$, $PC/CQ = 2/1$. Then $k=1/2$. Ratio = $\frac{(1/2+1)^2}{(1/2)^2 - (1/2) + 1} = \frac{(3/2)^2}{1/4 - 1/2 + 1} = \frac{9/4}{3/4} = 3$.

Let's assume the correct answer is indeed 5/2. This means $Area(\triangle ABC) = \frac{2}{5} Area(\triangle PQR)$.

Consider the formula: If points A, B, C divide sides QR, RP, PQ in ratios $m:n$, $p:q$, $r:s$ respectively. Then $Area(ABC)/Area(PQR) = 1 - \frac{m}{m+n}\frac{s}{r+s} - \frac{p}{p+q}\frac{n}{m+n} - \frac{r}{r+s}\frac{q}{p+q}$. Given $QA/AR = 1/2 \implies m=1, n=2$. $RB/BP = 1/2 \implies p=1, q=2$. $PC/CQ = 1/2 \implies r=1, s=2$.

$Area(ABC)/Area(PQR) = 1 - \frac{1}{3}\frac{2}{3} - \frac{1}{3}\frac{2}{3} - \frac{1}{3}\frac{2}{3} = 1 - \frac{2}{9} - \frac{2}{9} - \frac{2}{9} = 1 - \frac{6}{9} = \frac{1}{3}$. Ratio is 3.

Let's try to find a condition that yields 5/2. If $Area(ABC)/Area(PQR) = 2/5$. $1 - (\frac{m}{m+n}\frac{s}{r+s} + \frac{p}{p+q}\frac{n}{m+n} + \frac{r}{r+s}\frac{q}{p+q}) = 2/5$. $\frac{m}{m+n}\frac{s}{r+s} + \frac{p}{p+q}\frac{n}{m+n} + \frac{r}{r+s}\frac{q}{p+q} = 3/5$.

If $m/n = p/q = r/s = x$. Then $\frac{x}{(x+1)^2} + \frac{x}{(x+1)^2} + \frac{x}{(x+1)^2} = \frac{3x}{(x+1)^2} = 3/5$. $15x = 3(x+1)^2 = 3(x^2+2x+1)$. $5x = x^2+2x+1$. $x^2 - 3x + 1 = 0$. $x = \frac{3 \pm \sqrt{9-4}}{2} = \frac{3 \pm \sqrt{5}}{2}$. So if $QA/AR = RB/BP = PC/CQ = \frac{3+\sqrt{5}}{2}$ or $\frac{3-\sqrt{5}}{2}$, the ratio of areas would be 5/2. But the given ratio is 1/2.

Let's consider the formula for the area ratio when points divide the sides in ratio $m:n$. $Area(ABC)/Area(PQR) = \frac{m^2 - mn + n^2}{(m+n)^2}$. For $m=1, n=2$, we get $3/9 = 1/3$. Ratio is 3.

There might be a typo in the question or the given answer. However, if we assume the question meant that A, B, C divide the sides such that the remaining segment is twice the divided segment, i.e., $AR=2QA$, $BP=2RB$, $CQ=2PC$. This is what $QA/AR = 1/2$ implies.

Let's check the problem source or similar problems. A common problem states that points divide the sides in ratio $1:2$. Example: If A, B, C divide QR, RP, PQ in ratio $1:2$, then $Area(ABC)/Area(PQR) = 1/3$. Ratio is 3.

Let's consider an alternative interpretation: If $QA:QR = 1:2$, $RB:RP = 1:2$, $PC:PQ = 1:2$. Then $A$ is midpoint of QR, $B$ is midpoint of RP, $C$ is midpoint of PQ. Area ratio is 1/4. Ratio is 4.

If $QA:AR = 1:2$, $AR:RP = 1:2$, $RP:PQ = 1:2$. This is not possible.

Given the provided solution is A (5/2), let's try to reverse-engineer it. If $Area(\triangle PQR) / Area(\triangle ABC) = 5/2$. Then $Area(\triangle ABC) = \frac{2}{5} Area(\triangle PQR)$.

Consider the formula $Area(ABC)/Area(PQR) = 1 - (\frac{AR}{QR}\frac{BP}{RP}\frac{CQ}{PQ} + \frac{QA}{QR}\frac{RB}{RP}\frac{PC}{PQ})$. This is incorrect.

Let's assume the question implies $QA/QR = 1/2$, $RB/RP = 1/2$, $PC/PQ = 1/2$. This means A, B, C are midpoints. Area ratio is 1/4. Ratio is 4.

Let's assume the question implies $AR/QR = 1/2$, $BP/RP = 1/2$, $CQ/PQ = 1/2$. This means A, B, C are midpoints. Area ratio is 1/4. Ratio is 4.

Let's assume the question implies $QA/AR = 2/1$, $RB/BP = 2/1$, $PC/CQ = 2/1$. So $QA = 2AR$. $QR = 3AR$. $QA/QR = 2/3$. $RB = 2BP$. $RP = 3BP$. $RB/RP = 2/3$. $PC = 2CQ$. $PQ = 3CQ$. $PC/PQ = 2/3$. So A, B, C divide the sides in ratio $2:1$. Using the formula $Area(ABC)/Area(PQR) = \frac{m^2 - mn + n^2}{(m+n)^2}$ with $m=2, n=1$. $Area(ABC)/Area(PQR) = \frac{2^2 - 2(1) + 1^2}{(2+1)^2} = \frac{4 - 2 + 1}{3^2} = \frac{3}{9} = \frac{1}{3}$. Ratio is 3.

Let's consider the case where the ratios are applied in a cyclic manner but with different values that might lead to 5/2. If $QA/AR = 1/2$, $RB/BP = 1/3$, $PC/CQ = 1/4$. This is unlikely to be the case.

Given the consistency of the result 3 across multiple standard methods, and the discrepancy with the provided answer 5/2, there is a high probability of an error in the provided answer. However, if forced to choose an option that is not 3, and assuming a common exam pattern, there might be a subtle interpretation.

Let's assume the question meant: $AR/AQ = 1/2 \implies AQ = 2AR$. $QR = AQ+AR = 3AR$. $AR/QR = 1/3$. $BP/BR = 1/2 \implies BR = 2BP$. $RP = 3BP$. $BP/RP = 1/3$. $CQ/CP = 1/2 \implies CP = 2CQ$. $PQ = 3CQ$. $CQ/PQ = 1/3$. This means A, B, C divide the sides in ratio $2:1$. $QA/AR = 2/1$, $RB/BP = 2/1$, $PC/CQ = 2/1$. This leads to ratio 3 again.

Let's consider the possibility that the question meant: $QA/QR = 1/2$, $RB/RP = 1/2$, $PC/PQ = 1/2$. Midpoints. Ratio is 4.

Let's assume the question meant: $AR/QR = 1/2$, $BP/RP = 1/2$, $CQ/PQ = 1/2$. Midpoints. Ratio is 4.

If the answer is 5/2, then $Area(ABC) = \frac{2}{5} Area(PQR)$. Area of corner triangles sum to $3/5 Area(PQR)$.

Let's assume the ratios are $1:x$. The area ratio is $\frac{(x+1)^2}{x^2-x+1}$. If this is 5/2, then $\frac{(x+1)^2}{x^2-x+1} = 5/2$. $2(x^2+2x+1) = 5(x^2-x+1)$. $2x^2+4x+2 = 5x^2-5x+5$. $3x^2 - 9x + 3 = 0$. $x^2 - 3x + 1 = 0$. $x = \frac{3 \pm \sqrt{5}}{2}$. So if the ratio was $\frac{3 \pm \sqrt{5}}{2}$, the answer would be 5/2.

Given the consistent result of 3 from multiple standard methods for the given ratios, and the discrepancy with the provided answer, it is highly probable that the provided answer is incorrect. However, if we must select an answer from the options, and assuming there is no typo in the question, we are in a difficult situation.

Let's consider a scenario where the question meant that the lengths are in a ratio. $QA = k, AR = 2k$. $RB = k, BP = 2k$. $PC = k, CQ = 2k$.

Let's review the question and options one last time. The question is from JEE 2023. It's unlikely to have a simple error in the question itself. The only plausible explanation is a subtle interpretation of the ratios or a known formula that I am not recalling correctly.

Let's go back to the formula: $Area(ABC)/Area(PQR) = 1 - \frac{AR}{QR}\frac{BP}{RP}\frac{CQ}{PQ} - \frac{QA}{QR}\frac{RB}{RP}\frac{PC}{PQ}$. This formula is incorrect.

The correct formula for area of inner triangle ABC when A on QR, B on RP, C on PQ is: $Area(ABC) = Area(PQR) - Area(AQC) - Area(BPA) - Area(CRB)$. $QA/AR = 1/2 \implies QA/QR = 1/3, AR/QR = 2/3$. $RB/BP = 1/2 \implies RB/RP = 1/3, BP/RP = 2/3$. $PC/CQ = 1/2 \implies PC/PQ = 1/3, CQ/PQ = 2/3$.

$Area(\triangle AQC) = \frac{AQ}{QR} \cdot \frac{CQ}{PQ} Area(\triangle PQR)$ is incorrect. The correct formula is: $Area(\triangle CQA) = \frac{CQ}{PQ} \cdot \frac{QA}{QR} Area(\triangle PQR)$. $Area(\triangle CQA) = \frac{2}{3} \cdot \frac{1}{3} \Delta = \frac{2}{9} \Delta$.

$Area(\triangle APB) = \frac{AP}{RP} \cdot \frac{PB}{PQ} Area(\triangle PQR)$. Not directly applicable.

Let's use the formula: $Area(ABC) = Area(PQR) - Area(CQA) - Area(APB) - Area(BRC)$. $Area(CQA) = \frac{CQ}{PQ} \cdot \frac{QA}{QR} Area(PQR) = \frac{2}{3} \cdot \frac{1}{3} \Delta = \frac{2}{9} \Delta$. $Area(APB) = \frac{AP}{RP} \cdot \frac{PB}{PQ} Area(PQR)$.

Let's use the formula for areas of corner triangles: $Area(\triangle AQC) = \frac{AQ}{QR} \cdot \frac{QC}{QP} \sin(\angle RQP)$ is incorrect.

Let's use the property: If A, B, C divide QR, RP, PQ in ratio $1:k$. $Area(ABC)/Area(PQR) = \frac{k^2-k+1}{(k+1)^2}$. For $k=2$, this is $3/9=1/3$.

There might be a typo in the question, and it should have been $QA/AR = 2/1$, etc. If $QA/AR = 2/1$, then $k=1/2$. Ratio = $\frac{(1/2)^2 - (1/2) + 1}{(1/2+1)^2} = \frac{1/4 - 1/2 + 1}{(3/2)^2} = \frac{3/4}{9/4} = 1/3$. Ratio is 3.

Let's consider the formula $Area(ABC)/Area(PQR) = 1 - \frac{AR}{QR}\frac{BP}{RP}\frac{CQ}{PQ} - \frac{QA}{QR}\frac{RB}{RP}\frac{PC}{PQ}$ - this formula is incorrect.

Let's assume the correct answer is 5/2 and try to find a way to get it. If $Area(\triangle ABC) = \frac{2}{5} Area(\triangle PQR)$.

Let's consider the case where the ratios are $1:2$, $2:1$, $1:2$. If $QA/AR = 1/2$, $RB/BP = 2/1$, $PC/CQ = 1/2$. $\frac{m}{m+n} = 1/3$, $\frac{p}{p+q} = 2/3$, $\frac{r}{r+s} = 1/3$. $\frac{n}{m+n} = 2/3$, $\frac{q}{p+q} = 1/3$, $\frac{s}{r+s} = 2/3$.

$Area(ABC)/Area(PQR) = 1 - (\frac{1}{3}\frac{2}{3} + \frac{2}{3}\frac{2}{3} + \frac{1}{3}\frac{1}{3})$ $= 1 - (\frac{2}{9} + \frac{4}{9} + \frac{1}{9}) = 1 - \frac{7}{9} = \frac{2}{9}$. Ratio is 9/2.

If $QA/AR = 1/2$, $RB/BP = 1/2$, $PC/CQ = 2/1$. $Area(ABC)/Area(PQR) = 1 - (\frac{1}{3}\frac{1}{3} + \frac{1}{3}\frac{2}{3} + \frac{2}{3}\frac{1}{3})$ $= 1 - (\frac{1}{9} + \frac{2}{9} + \frac{2}{9}) = 1 - \frac{5}{9} = \frac{4}{9}$. Ratio is 9/4.

Given the strong evidence from standard formulas and vector methods that the ratio is 3, and the discrepancy with the provided answer, it is highly likely that the provided answer is incorrect. However, if a solution must be provided that matches the given correct answer, it suggests a non-standard interpretation or formula.

Let's assume there's a formula: $Area(ABC)/Area(PQR) = 1 - \frac{AR}{QR} \frac{BP}{RP} \frac{CQ}{PQ} - \frac{QA}{QR} \frac{RB}{RP} \frac{PC}{PQ}$. This is incorrect.

Let's consider the possibility of a typo in the question itself, e.g., the ratios are different. If $QA/AR = 1/x$, $RB/BP = 1/x$, $PC/CQ = 1/x$. Ratio = $\frac{(x+1)^2}{x^2-x+1}$. If this is 5/2, then $x = \frac{3 \pm \sqrt{5}}{2}$.

Given the problem is from JEE 2023, it's unlikely to have an error. Let's check if there's a special case of the formula.

Consider the formula: $Area(ABC)/Area(PQR) = \frac{(m^2-mn+n^2)}{(m+n)^2}$ for points dividing sides in ratio $m:n$. For $m=1, n=2$, $Area(ABC)/Area(PQR) = 3/9 = 1/3$. Ratio is 3.

Let's assume the question meant $AR/QA = 1/2$, $BP/RB = 1/2$, $CQ/PC = 1/2$. This means $QA=2AR$, $RB=2BP$, $PC=2CQ$. So $QA/QR = 2/3$, $RB/RP = 2/3$, $PC/PQ = 2/3$. Using $m=2, n=1$ in the formula $\frac{m^2-mn+n^2}{(m+n)^2} = \frac{2^2-2(1)+1^2}{(2+1)^2} = \frac{3}{9} = 1/3$. Ratio is 3.

It is possible that the provided "correct answer" is indeed incorrect. Based on standard geometric formulas and vector algebra, the ratio is consistently 3.

However, if we are forced to match the answer 5/2, let's consider a less common formula or interpretation. The problem asks for $Area(PQR)/Area(ABC)$. If the ratio is 5/2, then $Area(ABC) = \frac{2}{5} Area(PQR)$.

Let's assume the formula is of the form $1 - x$, where $x$ is the sum of areas of corner triangles. $x = \frac{AR}{QR}\frac{BP}{RP}\frac{CQ}{PQ} + \frac{QA}{QR}\frac{RB}{RP}\frac{PC}{PQ}$. This formula is incorrect.

Given the situation, and the high probability of error in the provided answer, I will present the solution that consistently yields 3. However, if the provided answer of 5/2 is strictly to be matched, then the problem requires a different approach or formula not readily apparent.

Let's reconsider the vector calculation. $\vec{a} = \frac{2\vec{q} + \vec{r}}{3}$ $\vec{b} = \frac{2\vec{r} + \vec{p}}{3}$ $\vec{c} = \frac{2\vec{p} + \vec{q}}{3}$

$\vec{AB} = \vec{b} - \vec{a} = \frac{\vec{p} - 2\vec{q} + \vec{r}}{3}$ $\vec{AC} = \vec{c} - \vec{a} = \frac{2\vec{p} - \vec{q} - \vec{r}}{3}$

$Area(\triangle ABC) = \frac{1}{2} |(\vec{b}-\vec{a}) \times (\vec{c}-\vec{a})| = \frac{1}{18} |(\vec{p} - 2\vec{q} + \vec{r}) \times (2\vec{p} - \vec{q} - \vec{r})|$ $= \frac{1}{18} | 2\vec{p}\times\vec{p} - \vec{p}\times\vec{q} - \vec{p}\times\vec{r} - 4\vec{q}\times\vec{p} + 2\vec{q}\times\vec{q} + 2\vec{q}\times\vec{r} + 2\vec{r}\times\vec{p} - \vec{r}\times\vec{q} - \vec{r}\times\vec{r} |$ $= \frac{1}{18} | \vec{0} - \vec{p}\times\vec{q} - \vec{p}\times\vec{r} + 4\vec{p}\times\vec{q} + \vec{0} + 2\vec{q}\times\vec{r} + 2\vec{r}\times\vec{p} - \vec{r}\times\vec{q} + \vec{0} |$ $= \frac{1}{18} | 3\vec{p}\times\vec{q} - \vec{p}\times\vec{r} + 2\vec{q}\times\vec{r} + 2\vec{r}\times\vec{p} + \vec{q}\times\vec{r} |$ $= \frac{1}{18} | 3\vec{p}\times\vec{q} + \vec{p}\times\vec{r} + 3\vec{q}\times\vec{r} |$ $= \frac{1}{18} | 3(\vec{p}\times\vec{q} + \vec{q}\times\vec{r}) + \vec{p}\times\vec{r} |$ $= \frac{1}{18} | 3(\vec{p}\times\vec{q} + \vec{q}\times\vec{r} + \vec{r}\times\vec{p}) |$ $= \frac{1}{6} |\vec{p}\times\vec{q} + \vec{q}\times\vec{r} + \vec{r}\times\vec{p}|$.

$Area(\triangle PQR) = \frac{1}{2} |\vec{p}\times\vec{q} + \vec{q}\times\vec{r} + \vec{r}\times\vec{p}|$. So $Area(\triangle ABC) = \frac{1}{3} Area(\triangle PQR)$. Ratio is 3.

The problem states the correct answer is A (5/2). If the ratio $Area(\triangle PQR) / Area(\triangle ABC) = 5/2$, then $Area(\triangle ABC) = \frac{2}{5} Area(\triangle PQR)$. This implies that the sum of areas of the corner triangles is $3/5 Area(\triangle PQR)$.

Let's check if there's a formula of the form: $Area(ABC)/Area(PQR) = 1 - \sum (\text{ratios of sides})$.

Given the discrepancy, and the consistent result of 3 from standard methods, it is highly probable that the provided answer is incorrect. However, since I am instructed to provide a solution that matches the correct answer, and I cannot derive 5/2 from the given information using standard methods, I cannot fulfill that specific instruction without making unsubstantiated assumptions or using incorrect formulas.

Assuming the question and answer are correct, there must be a known theorem or a specific interpretation that leads to 5/2. Without further information or clarification, it is impossible to derive it.

Final attempt to find an interpretation that yields 5/2. If the ratios were $QA/AR = 1/1$, $RB/BP = 1/1$, $PC/CQ = 1/1$. (Midpoints) Ratio is 4.

If the ratios were $QA/AR = 1/3$, $RB/BP = 1/3$, $PC/CQ = 1/3$. $k=3$. Ratio = $\frac{(3+1)^2}{3^2-3+1} = \frac{16}{9-3+1} = \frac{16}{7}$.

If the question implied something like: $QA = 1, AR = 2$. $RB = 1, BP = 2$. $PC = 1, CQ = 2$. This is the same as ratio 1:2.

Given the context of a JEE exam, there is a standard formula for this type of problem. If points A, B, C divide the sides QR, RP, PQ in the ratio $m:n$, $p:q$, $r:s$ respectively, then $\frac{Area(ABC)}{Area(PQR)} = 1 - (\frac{m}{m+n}\frac{s}{r+s} + \frac{p}{p+q}\frac{n}{m+n} + \frac{r}{r+s}\frac{q}{p+q})$. Here $QA/AR = 1/2 \implies m=1, n=2$. $RB/BP = 1/2 \implies p=1, q=2$. $PC/CQ = 1/2 \implies r=1, s=2$.

$\frac{Area(ABC)}{Area(PQR)} = 1 - (\frac{1}{3}\frac{2}{3} + \frac{1}{3}\frac{2}{3} + \frac{1}{3}\frac{2}{3}) = 1 - (\frac{2}{9} + \frac{2}{9} + \frac{2}{9}) = 1 - \frac{6}{9} = 1 - \frac{2}{3} = \frac{1}{3}$. So the ratio $Area(PQR)/Area(ABC) = 3$.

Since I cannot derive the answer 5/2, and my derivation consistently leads to 3, I will present the solution for 3, noting the discrepancy.

Step-by-Step Solution (Revised to present the consistent result)

Step 1: Define position vectors for the vertices of $\triangle PQR$. Let the position vectors of points $P$, $Q$, and $R$ be $\vec{p}$, $\vec{q}$, and $\vec{r}$ respectively, with respect to an arbitrary origin $O$.

Step 2: Express the position vectors of points A, B, and C using the section formula. We are given the ratios: $\frac{QA}{AR} = \frac{1}{2}$. This means point $A$ divides the line segment $QR$ in the ratio $1:2$. Using the section formula, the position vector of $A$, $\vec{a}$, is: $\vec{a} = \frac{2\vec{q} + 1\vec{r}}{1+2} = \frac{2\vec{q} + \vec{r}}{3}$ $\frac{RB}{BP} = \frac{1}{2}$. This means point $B$ divides the line segment $RP$ in the ratio $1:2$. The position vector of $B$, $\vec{b}$, is: $\vec{b} = \frac{2\vec{r} + 1\vec{p}}{1+2} = \frac{2\vec{r} + \vec{p}}{3}$ $\frac{PC}{CQ} = \frac{1}{2}$. This means point $C$ divides the line segment $PQ$ in the ratio $1:2$. The position vector of $C$, $\vec{c}$, is: $\vec{c} = \frac{2\vec{p} + 1\vec{q}}{1+2} = \frac{2\vec{p} + \vec{q}}{3}$

Step 3: Calculate the area of $\triangle PQR$ in terms of position vectors. The area of a triangle with vertices having position vectors $\vec{p}$, $\vec{q}$, and $\vec{r}$ is given by $\frac{1}{2} |(\vec{q}-\vec{p}) \times (\vec{r}-\vec{p})|$. Expanding this, we get: $Area(\triangle PQR) = \frac{1}{2} |\vec{q}\times\vec{r} - \vec{q}\times\vec{p} - \vec{p}\times\vec{r} + \vec{p}\times\vec{p}|$ $Area(\triangle PQR) = \frac{1}{2} |\vec{q}\times\vec{r} + \vec{p}\times\vec{q} + \vec{r}\times\vec{p}|$ (Using $\vec{a}\times\vec{b} = -\vec{b}\times\vec{a}$ and $\vec{a}\times\vec{a} = \vec{0}$)

Step 4: Calculate the area of $\triangle ABC$ in terms of position vectors. The area of $\triangle ABC$ is given by $\frac{1}{2} |(\vec{b}-\vec{a}) \times (\vec{c}-\vec{a})|$. First, find the vectors representing the sides of $\triangle ABC$: $\vec{b}-\vec{a} = \left(\frac{2\vec{r} + \vec{p}}{3}\right) - \left(\frac{2\vec{q} + \vec{r}}{3}\right) = \frac{\vec{p} - 2\vec{q} + \vec{r}}{3}$ $\vec{c}-\vec{a} = \left(\frac{2\vec{p} + \vec{q}}{3}\right) - \left(\frac{2\vec{q} + \vec{r}}{3}\right) = \frac{2\vec{p} - \vec{q} - \vec{r}}{3}$ Now, calculate the cross product: $(\vec{b}-\vec{a}) \times (\vec{c}-\vec{a}) = \frac{1}{9} (\vec{p} - 2\vec{q} + \vec{r}) \times (2\vec{p} - \vec{q} - \vec{r})$ Expanding this cross product and using properties of vector cross products ($\vec{a}\times\vec{a}=\vec{0}$, $\vec{a}\times\vec{b}=-\vec{b}\times\vec{a}$): $= \frac{1}{9} [ 2\vec{p}\times\vec{p} - \vec{p}\times\vec{q} - \vec{p}\times\vec{r} - 4\vec{q}\times\vec{p} + 2\vec{q}\times\vec{q} + 2\vec{q}\times\vec{r} + 2\vec{r}\times\vec{p} - \vec{r}\times\vec{q} - \vec{r}\times\vec{r} ]$ $= \frac{1}{9} [ \vec{0} - \vec{p}\times\vec{q} - \vec{p}\times\vec{r} + 4\vec{p}\times\vec{q} + \vec{0} + 2\vec{q}\times\vec{r} + 2\vec{r}\times\vec{p} + \vec{q}\times\vec{r} + \vec{0} ]$ $= \frac{1}{9} [ 3\vec{p}\times\vec{q} + \vec{p}\times\vec{r} + 3\vec{q}\times\vec{r} ]$ $= \frac{1}{9} [ 3(\vec{p}\times\vec{q} + \vec{q}\times\vec{r}) + \vec{p}\times\vec{r} ]$ $= \frac{1}{9} [ 3(\vec{p}\times\vec{q} + \vec{q}\times\vec{r} + \vec{r}\times\vec{p}) ]$ $= \frac{1}{3} (\vec{p}\times\vec{q} + \vec{q}\times\vec{r} + \vec{r}\times\vec{p})$ So, the area of $\triangle ABC$ is: $Area(\triangle ABC) = \frac{1}{2} \left| \frac{1}{3} (\vec{p}\times\vec{q} + \vec{q}\times\vec{r} + \vec{r}\times\vec{p}) \right| = \frac{1}{3} \left( \frac{1}{2} |\vec{p}\times\vec{q} + \vec{q}\times\vec{r} + \vec{r}\times\vec{p}| \right)$ $Area(\triangle ABC) = \frac{1}{3} Area(\triangle PQR)$

Step 5: Calculate the ratio of the areas. We need to find the ratio $\frac{Area(\triangle PQR)}{Area(\triangle ABC)}$. $\frac{Area(\triangle PQR)}{Area(\triangle ABC)} = \frac{Area(\triangle PQR)}{\frac{1}{3} Area(\triangle PQR)} = 3$

Common Mistakes & Tips

• Misinterpreting Ratios: Ensure that the ratios given (e.g., $QA/AR = 1/2$) are correctly used in the section formula. $m:n$ means the point divides the segment into $m$ parts and $n$ parts.
• Vector Cross Product Properties: Be careful with the sign conventions and properties of the cross product, especially when expanding terms. Remember $\vec{a} \times \vec{b} = -\vec{b} \times \vec{a}$ and $\vec{a} \times \vec{a} = \vec{0}$.
• Area Formula: The area of a triangle with vertices $\vec{p}, \vec{q}, \vec{r}$ is $\frac{1}{2} |(\vec{q}-\vec{p}) \times (\vec{r}-\vec{p})|$, not just the magnitude of the cross product.

Summary

We used vector algebra and the section formula to find the position vectors of the vertices of the inner triangle $\triangle ABC$ in terms of the position vectors of the outer triangle $\triangle PQR$. By calculating the areas of both triangles using their position vectors and their cross products, we found that the area of $\triangle ABC$ is exactly one-third of the area of $\triangle PQR$. Therefore, the ratio of the area of $\triangle PQR$ to the area of $\triangle ABC$ is 3.

The final answer is \boxed{3}.

Note: The provided correct answer in the problem statement is (A) 5/2. However, standard geometric formulas and vector algebra consistently yield a ratio of 3 for the given ratios. This suggests a potential error in the provided correct answer. The derivation above leads to the result 3. If the answer is indeed 5/2, it would require a different interpretation of the ratios or a non-standard formula.

The final answer is \boxed{3}.