Square of Distance from Point to Line Measured Along Another Line

To find the square of the distance of the point $P(-2, -8, 6)$ from the line $L_1$ measured along the line $L_2$, we follow these steps:

1. Equation of the line $L_1$: $L_1: \frac{x-1}{1} = \frac{y-1}{2} = \frac{z}{-1}$

2. Direction ratios of the line $L_2$: The direction ratios of the line $L_2: \frac{x+5}{1} = \frac{y+5}{-1} = \frac{z}{2}$ are $(1, -1, 2)$.

3. Parametric representation of a point $Q$ on the line passing through $P$ parallel to $L_2$: Any point $Q$ on the line passing through $P(-2, -8, 6)$ and parallel to $L_2$ can be expressed in terms of a parameter $\lambda$ as: $Q = (-2 + \lambda, -8 - \lambda, 6 + 2\lambda)$

4. Finding the intersection point $Q$ on line $L_1$: Since $Q$ lies on $L_1$, its coordinates must satisfy the equation of $L_1$: $\frac{(-2 + \lambda) - 1}{1} = \frac{(-8 - \lambda) - 1}{2} = \frac{6 + 2\lambda}{-1}$

$\frac{\lambda - 3}{1} = \frac{-\lambda - 9}{2} = \frac{-2\lambda - 6}{1}$

Taking the first two ratios: $2(\lambda - 3) = -\lambda - 9$ $2\lambda - 6 = -\lambda - 9$ $3\lambda = -3 \implies \lambda = -1$

Checking with the third ratio: $\frac{\lambda - 3}{1} = \frac{-1 - 3}{1} = -4$ $\frac{-2\lambda - 6}{1} = \frac{-2(-1) - 6}{1} = -4$

Thus, $\lambda = -1$ is consistent.

5. Calculating the square of the distance $PQ$: The coordinates of $Q$ are: $Q = (-2 - 1, -8 - (-1), 6 + 2(-1)) = (-3, -7, 4)$

The square of the distance $PQ^2$ is given by: $PQ^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$ $PQ^2 = (-3 - (-2))^2 + (-7 - (-8))^2 + (4 - 6)^2$ $PQ^2 = (-1)^2 + (1)^2 + (-2)^2 = 1 + 1 + 4 = 6$

Hence, the square of the distance is 6.

Correct Option: B