1. Key Concepts and Formulas

• Equation of a Line in 3D: A line passing through a point $A(x_0, y_0, z_0)$ and parallel to a direction vector $\vec{d} = a\hat{i}+b\hat{j}+c\hat{k}$ can be represented parametrically as $(x_0+at, y_0+bt, z_0+ct)$, where $t$ is a scalar parameter. This allows us to express any point on the line in terms of $t$.
• Distance Formula in 3D: The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.
• Area of a Triangle in 3D: If two vectors $\vec{u}$ and $\vec{v}$ represent two sides of a triangle originating from a common vertex, the area of the triangle is given by $\frac{1}{2} \|\vec{u} \times \vec{v}\|$. The square of the area is then $\frac{1}{4} \|\vec{u} \times \vec{v}\|^2$.

2. Step-by-Step Solution

Step 1: Represent Point Q in Parametric Form

• What we are doing: We are defining the coordinates of point $Q$ using the given information that it lies on a line passing through $P$ and parallel to a specific vector.
• Why we are doing it: This allows us to express the unknown coordinates of $Q$ in terms of a single variable, $t$, which can then be solved for using the distance condition. Given point $P(-2,-1,3)$ and direction vector $\vec{d} = 3\hat{i}+2\hat{j}+2\hat{k}$. Any point $Q$ on the line passing through $P$ and parallel to $\vec{d}$ can be written as: $Q = P + t\vec{d}$ $Q(x_Q, y_Q, z_Q) = (-2 + 3t,\, -1 + 2t,\, 3 + 2t)$ The problem states that $P$ and $Q$ are distinct points, which implies $t \neq 0$.

Step 2: Use the Distance Condition to Find the Parameter 't'

• What we are doing: We are using the given distance between $Q$ and $R$ to form an equation and solve for the parameter $t$.
• Why we are doing it: Finding $t$ will allow us to determine the exact coordinates of point $Q$. We are given that the distance $QR = 5$. Point $R$ is $(1,3,3)$. Using the distance formula, $QR^2 = 5^2 = 25$: $(x_Q - x_R)^2 + (y_Q - y_R)^2 + (z_Q - z_R)^2 = 25$ Substitute the coordinates of $Q$ and $R$: $((-2 + 3t) - 1)^2 + ((-1 + 2t) - 3)^2 + ((3 + 2t) - 3)^2 = 25$ Simplify the terms inside the parentheses: $(3t - 3)^2 + (2t - 4)^2 + (2t)^2 = 25$ Expand and simplify the equation: $9(t^2 - 2t + 1) + 4(t^2 - 4t + 4) + 4t^2 = 25$ $9t^2 - 18t + 9 + 4t^2 - 16t + 16 + 4t^2 = 25$ Combine like terms: $(9t^2 + 4t^2 + 4t^2) + (-18t - 16t) + (9 + 16) = 25$ $17t^2 - 34t + 25 = 25$ Subtract 25 from both sides: $17t^2 - 34t = 0$ Factor out $17t$: $17t(t - 2) = 0$ This gives two possible values for $t$: $t=0$ or $t=2$. Since $P$ and $Q$ are distinct points, $t \neq 0$. Therefore, $t=2$.

Step 3: Determine the Coordinates of Point Q

• What we are doing: We are substituting the value of $t$ back into the parametric form of $Q$ to find its exact coordinates.
• Why we are doing it: We need the coordinates of $Q$ to form the vectors required for calculating the triangle's area. Substitute $t=2$ into $Q = (-2 + 3t,\, -1 + 2t,\, 3 + 2t)$: $Q = (-2 + 3(2),\, -1 + 2(2),\, 3 + 2(2))$ $Q = (-2 + 6,\, -1 + 4,\, 3 + 4)$ $Q = (4, 3, 7)$ So, the coordinates of point $Q$ are $(4,3,7)$.

Step 4: Calculate Vectors $\vec{PQ}$ and $\vec{PR}$

• What we are doing: We are forming two vectors that represent two sides of the triangle $\triangle PQR$ originating from a common vertex $P$.
• Why we are doing it: These vectors are necessary for calculating the cross product, which is used in the area formula. Given $P(-2,-1,3)$, $Q(4,3,7)$, and $R(1,3,3)$. Vector $\vec{PQ}$: $\vec{PQ} = Q - P = (4 - (-2),\, 3 - (-1),\, 7 - 3) = (6, 4, 4)$ Vector $\vec{PR}$: $\vec{PR} = R - P = (1 - (-2),\, 3 - (-1),\, 3 - 3) = (3, 4, 0)$

Step 5: Compute the Cross Product $\vec{PQ} \times \vec{PR}$

• What we are doing: We are calculating the cross product of the two vectors $\vec{PQ}$ and $\vec{PR}$.
• Why we are doing it: The magnitude of this cross product is directly related to the area of the parallelogram formed by the vectors, and thus to the area of the triangle. $\vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 4 & 4 \\ 3 & 4 & 0 \end{vmatrix}$ Expand the determinant:
  • $\hat{i}$-component: $(4 \times 0) - (4 \times 4) = 0 - 16 = -16$
  • $\hat{j}$-component: $-((6 \times 0) - (4 \times 3)) = -(0 - 12) = 12$
  • $\hat{k}$-component: $(6 \times 4) - (4 \times 3) = 24 - 12 = 12$ So, the cross product vector is: $\vec{PQ} \times \vec{PR} = -16\hat{i} + 12\hat{j} + 12\hat{k}$

Step 6: Calculate the Square of the Area of $\triangle PQR$

• What we are doing: We are using the formula for the square of the area of a triangle, which involves the magnitude squared of the cross product.
• Why we are doing it: This is the final quantity requested by the problem. The square of the area of $\triangle PQR$ is given by $\frac{1}{4} \|\vec{PQ} \times \vec{PR}\|^2$. First, calculate the magnitude squared of the cross product vector: $\|\vec{PQ} \times \vec{PR}\|^2 = (-16)^2 + (12)^2 + (12)^2$ $= 256 + 144 + 144$ $= 544$ Now, calculate the square of the area: $\text{Area}^2 = \frac{1}{4} \times 544 = 136$

3. Common Mistakes & Tips

• Forgetting $t \neq 0$: Always check conditions like "distinct points" as they can eliminate extraneous solutions for the parameter $t$.
• Cross Product Sign Error: Remember the negative sign for the $\hat{j}$ component when expanding a $3 \times 3$ determinant for the cross product.
• Magnitude vs. Magnitude Squared: Pay close attention to whether the problem asks for the area or the square of the area, and calculate $\|\vec{u} \times \vec{v}\|$ or $\|\vec{u} \times \vec{v}\|^2$ accordingly.

4. Summary

We first expressed point $Q$ parametrically using point $P$ and the given direction vector. Then, we used the distance condition between $Q$ and $R$ to solve for the parameter $t$, which allowed us to find the exact coordinates of $Q$. With the coordinates of $P$, $Q$, and $R$, we formed vectors $\vec{PQ}$ and $\vec{PR}$. Finally, we computed their cross product, found its magnitude squared, and divided by 4 to obtain the square of the area of $\triangle PQR$. The calculated square of the area is 136.

5. Final Answer

The square of the area of $\triangle PQR$ is 136.

The final answer is $\boxed{\text{136}}$, which corresponds to option (C).