The square of the distance of the point ($-2$, $-8$, 6) from the line $\frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z}{-1}$ along the line $\frac{x + 5}{1} = \frac{y + 5}{-1} = \frac{z}{2}$ is equal to:

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The square of the distance of the point ($-2$, $-8$, 6) from the line $\frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z}{-1}$ along the line $\frac{x + 5}{1} = \frac{y + 5}{-1} = \frac{z}{2}$ is equal to:

The correct answer is

The problem asks for the square of the distance of the point P($-2$, $-8$, 6) from the line $L_1: \frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z}{-1}$. The distance is measured "along the line" $L_2: \frac{x + 5}{1} = \frac{y + 5}{-1} = \frac{z}{2}$. This means we need to find a point Q on $L_1$ such that the vector $\vec{PQ}$ is parallel to the direction vector of $L_2$.

1. Define Point and Lines

Given Point: P = ($-2$, $-8$, 6).
Line $L_1$: Passes through A(1, 1, 0) with direction vector $\vec{d_1} = <1, 2, -1>$.
Line $L_2$: Has direction vector $\vec{d_2} = <1, -1, 2>$.

2. Parametric Representation of Line 1

Any point Q on line $L_1$ can be expressed parametrically using the parameter $t$:

$Q = (1 + 1 \cdot t, 1 + 2 \cdot t, 0 + (-1) \cdot t) = (1 + t, 1 + 2t, -t)$

3. Vector from Point P to Point Q

Calculate the vector $\vec{PQ}$ using the coordinates of P and Q:

$\vec{PQ} = Q - P = <(1 + t) - (-2), (1 + 2t) - (-8), (-t) - 6>$

$\vec{PQ} = <3 + t, 9 + 2t, -6 - t>$

4. Condition for Vector PQ along Line 2

For $\vec{PQ}$ to be "along the line" $L_2$, it must be parallel to the direction vector $\vec{d_2}$. Therefore, $\vec{PQ} = k \cdot \vec{d_2}$ for some scalar $k$.

$<3 + t, 9 + 2t, -6 - t> = k <1, -1, 2>$

Equating the components gives a system of linear equations:

$3 + t = k$
$9 + 2t = -k$
$-6 - t = 2k$

5. Solve for Parameter t

Substitute $k$ from the first equation into the second equation:

$9 + 2t = -(3 + t)$

$9 + 2t = -3 - t$

Combine terms involving $t$ and constants:

$3t = -12$

$t = -4$

(This value of $t$ is consistent with the third equation as well.)

6. Determine Point Q

Substitute $t = -4$ back into the parametric equation for Q:

$Q = (1 + (-4), 1 + 2(-4), -(-4)) = (-3, -7, 4)$

7. Calculate Vector PQ and its Magnitude Squared

Find the vector $\vec{PQ}$ using $t=-4$ (or coordinates of P and Q):

$\vec{PQ} = <3 + (-4), 9 + 2(-4), -6 - (-4)> = <-1, 1, -2>$

The square of the distance is the square of the magnitude (length) of $\vec{PQ}$:

$Distance^2 = ||\vec{PQ}||^2 = (-1)^2 + (1)^2 + (-2)^2$

$Distance^2 = 1 + 1 + 4 = 6$

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