1. Key Concepts and Formulas

• Foot of the Perpendicular from a Point to a Line: To find the foot of the perpendicular $Q$ from a point $P_0(x_0, y_0, z_0)$ to a line $L$ given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$:

  1. Represent a general point $Q$ on the line $L$ parametrically, say $Q(x_1 + a\lambda, y_1 + b\lambda, z_1 + c\lambda)$.
  2. Form the vector $\vec{P_0Q}$.
  3. The direction vector of the line is $\vec{d} = a\hat{i} + b\hat{j} + c\hat{k}$.
  4. Since $\vec{P_0Q}$ is perpendicular to $L$, it must be perpendicular to $\vec{d}$. Thus, their dot product is zero: $\vec{P_0Q} \cdot \vec{d} = 0$.
  5. Solve for $\lambda$ and substitute it back into the parametric coordinates of $Q$ to find the foot of the perpendicular.

• Area of a Triangle using Vector Cross Product: Given three vertices $P, Q, R$ of a triangle, its area $A$ can be calculated as half the magnitude of the cross product of two vectors representing two sides of the triangle originating from a common vertex (e.g., $\vec{PQ}$ and $\vec{PR}$): $A = \frac{1}{2} |\vec{PQ} \times \vec{PR}|$

2. Step-by-Step Solution

Step 1: Understand the given information and goal. We are given a point $P(5, 1, -3)$ and two lines $L_1: x - 1 = y - 2 = z$ and $L_2: x - 2 = y = z - 1$. Our objective is to:

1. Find the coordinates of $Q$, the foot of the perpendicular from $P$ to $L_1$.
2. Find the coordinates of $R$, the foot of the perpendicular from $P$ to $L_2$.
3. Calculate the area $A$ of the triangle $PQR$.
4. Finally, determine the value of $4A^2$.

Step 2: Find the coordinates of Q, the foot of the perpendicular from P to L1.

• 2.1: Express $L_1$ in symmetric and parametric form and identify its direction vector. The line $L_1$ is given by $x - 1 = y - 2 = z$. We can write this in standard symmetric form as $\frac{x-1}{1} = \frac{y-2}{1} = \frac{z-0}{1}$. This form indicates that $L_1$ passes through the point $(1, 2, 0)$ and has a direction vector $\vec{d_1} = \hat{i} + \hat{j} + \hat{k}$.
• 2.2: Represent a general point Q on $L_1$ parametrically. Let $Q$ be any point on $L_1$. Its coordinates can be expressed using a parameter $\lambda_1$: $Q(1 + \lambda_1, 2 + \lambda_1, \lambda_1)$
• 2.3: Form the vector $\vec{PQ}$. The vector connecting $P(5, 1, -3)$ to $Q(1 + \lambda_1, 2 + \lambda_1, \lambda_1)$ is: $\vec{PQ} = ((1 + \lambda_1) - 5)\hat{i} + ((2 + \lambda_1) - 1)\hat{j} + (\lambda_1 - (-3))\hat{k}$ $\vec{PQ} = (\lambda_1 - 4)\hat{i} + (\lambda_1 + 1)\hat{j} + (\lambda_1 + 3)\hat{k}$
• 2.4: Apply the perpendicularity condition to find $\lambda_1$. Since $Q$ is the foot of the perpendicular, the vector $\vec{PQ}$ must be perpendicular to the direction vector of $L_1$, $\vec{d_1}$. Their dot product must be zero: $\vec{PQ} \cdot \vec{d_1} = 0$ $(\lambda_1 - 4)(1) + (\lambda_1 + 1)(1) + (\lambda_1 + 3)(1) = 0$ $\lambda_1 - 4 + \lambda_1 + 1 + \lambda_1 + 3 = 0$ $3\lambda_1 = 0 \implies \lambda_1 = 0$
• 2.5: Substitute $\lambda_1$ to find the coordinates of Q. Substitute $\lambda_1 = 0$ back into the parametric coordinates of $Q$: $Q(1 + 0, 2 + 0, 0) \implies Q(1, 2, 0)$ So, the foot of the perpendicular from $P$ to $L_1$ is $Q(1, 2, 0)$.

Step 3: Find the coordinates of R, the foot of the perpendicular from P to L2.

• 3.1: Express $L_2$ in symmetric and parametric form and identify its direction vector. The line $L_2$ is given by $x - 2 = y = z - 1$. We can write this in standard symmetric form as $\frac{x-2}{1} = \frac{y-0}{1} = \frac{z-1}{1}$. This form indicates that $L_2$ passes through the point $(2, 0, 1)$ and has a direction vector $\vec{d_2} = \hat{i} + \hat{j} + \hat{k}$.
• 3.2: Represent a general point R on $L_2$ parametrically. Let $R$ be any point on $L_2$. Its coordinates can be expressed using a parameter $\lambda_2$: $R(2 + \lambda_2, \lambda_2, 1 + \lambda_2)$
• 3.3: Form the vector $\vec{PR}$. The vector connecting $P(5, 1, -3)$ to $R(2 + \lambda_2, \lambda_2, 1 + \lambda_2)$ is: $\vec{PR} = ((2 + \lambda_2) - 5)\hat{i} + (\lambda_2 - 1)\hat{j} + ((1 + \lambda_2) - (-3))\hat{k}$ $\vec{PR} = (\lambda_2 - 3)\hat{i} + (\lambda_2 - 1)\hat{j} + (\lambda_2 + 4)\hat{k}$
• 3.4: Apply the perpendicularity condition to find $\lambda_2$. Since $R$ is the foot of the perpendicular, the vector $\vec{PR}$ must be perpendicular to the direction vector of $L_2$, $\vec{d_2}$. Their dot product must be zero: $\vec{PR} \cdot \vec{d_2} = 0$ $(\lambda_2 - 3)(1) + (\lambda_2 - 1)(1) + (\lambda_2 + 4)(1) = 0$ $\lambda_2 - 3 + \lambda_2 - 1 + \lambda_2 + 4 = 0$ $3\lambda_2 = 0 \implies \lambda_2 = 0$
• 3.5: Substitute $\lambda_2$ to find the coordinates of R. Substitute $\lambda_2 = 0$ back into the parametric coordinates of $R$: $R(2 + 0, 0, 1 + 0) \implies R(2, 0, 1)$ So, the foot of the perpendicular from $P$ to $L_2$ is $R(2, 0, 1)$.

Step 4: Calculate the area A of triangle PQR.

• 4.1: List the coordinates of the vertices P, Q, R. We have $P(5, 1, -3)$, $Q(1, 2, 0)$, and $R(2, 0, 1)$.
• 4.2: Form the vectors $\vec{PQ}$ and $\vec{PR}$. $\vec{PQ} = Q - P = (1-5)\hat{i} + (2-1)\hat{j} + (0-(-3))\hat{k} = -4\hat{i} + \hat{j} + 3\hat{k}$ $\vec{PR} = R - P = (2-5)\hat{i} + (0-1)\hat{j} + (1-(-3))\hat{k} = -3\hat{i} - \hat{j} + 4\hat{k}$
• 4.3: Calculate the cross product $\vec{PQ} \times \vec{PR}$. The cross product is computed as the determinant: $\vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & 1 & 3 \\ -3 & -1 & 4 \end{vmatrix}$ $= \hat{i}((1)(4) - (3)(-1)) - \hat{j}((-4)(4) - (3)(-3)) + \hat{k}((-4)(-1) - (1)(-3))$ $= \hat{i}(4 + 3) - \hat{j}(-16 + 9) + \hat{k}(4 + 3)$ $= 7\hat{i} - \hat{j}(-7) + 7\hat{k}$ $= 7\hat{i} + 7\hat{j} + 7\hat{k}$
• 4.4: Calculate the magnitude of the cross product. $|\vec{PQ} \times \vec{PR}| = |7\hat{i} + 7\hat{j} + 7\hat{k}| = \sqrt{7^2 + 7^2 + 7^2}$ $= \sqrt{49 + 49 + 49} = \sqrt{3 \times 49} = 7\sqrt{3}$
• 4.5: Calculate the area A of triangle PQR. The area $A$ is half the magnitude of the cross product: $A = \frac{1}{2} |\vec{PQ} \times \vec{PR}| = \frac{1}{2} (7\sqrt{3}) = \frac{7\sqrt{3}}{2}$

Step 5: Calculate $4A^2$.

• 5.1: Square the area A. $A^2 = \left(\frac{7\sqrt{3}}{2}\right)^2 = \frac{7^2 \times (\sqrt{3})^2}{2^2} = \frac{49 \times 3}{4} = \frac{147}{4}$
• 5.2: Multiply by 4. $4A^2 = 4 \times \frac{147}{4} = 147$

3. Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when calculating vector components (e.g., $x_B - x_A$) and especially during the cross product determinant expansion. A single sign error can propagate and lead to an incorrect final answer.
• Parametric Representation: Ensure you correctly convert the line equation into parametric form. Each line gets its own independent parameter ($\lambda_1$, $\lambda_2$).
• Dot Product for Perpendicularity: Remember that the dot product of two perpendicular vectors is zero. This is the fundamental condition for finding the foot of the perpendicular.
• Cross Product Calculation: Practice calculating the $3 \times 3$ determinant for the cross product to avoid computational errors.

4. Summary

We first found the feet of the perpendiculars, $Q(1, 2, 0)$ and $R(2, 0, 1)$, by representing general points on lines $L_1$ and $L_2$ parametrically and using the condition that the vector from $P$ to the foot of the perpendicular must be orthogonal to the line's direction vector. Then, we formed vectors $\vec{PQ}$ and $\vec{PR}$ and calculated their cross product. The magnitude of this cross product, divided by two, gave us the area $A = \frac{7\sqrt{3}}{2}$. Finally, we computed $4A^2 = 147$.

5. Final Answer

The final answer is $\boxed{\text{147}}$, which corresponds to option (A).