The square of the distance of the point (-2, -8, 6) from the line...

Let the given point be $P(-2, -8, 6)$.

The distance is measured along the line $\dfrac{x+5}{1} = \dfrac{y+5}{-1} = \dfrac{z}{2}$, which has direction ratios $(1, -1, 2)$.

The equation of the line passing through $P$ and parallel to this line is:
$\dfrac{x+2}{1} = \dfrac{y+8}{-1} = \dfrac{z-6}{2} = \lambda$

Any point on this line can be written as $Q(\lambda - 2, -\lambda - 8, 2\lambda + 6)$.

Since $Q$ lies on the line $\dfrac{x-1}{1} = \dfrac{y-1}{2} = \dfrac{z}{-1}$, we substitute the coordinates of $Q$ into its equation:
$\dfrac{\lambda - 2 - 1}{1} = \dfrac{-\lambda - 8 - 1}{2} = \dfrac{2\lambda + 6}{-1}$

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

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

Substituting $\lambda = -1$ into the coordinates of $Q$, we get:
$Q(-1 - 2, -(-1) - 8, 2(-1) + 6) \equiv Q(-3, -7, 4)$

The square of the distance between $P(-2, -8, 6)$ and $Q(-3, -7, 4)$ is:
$PQ^2 = (-3 - (-2))^2 + (-7 - (-8))^2 + (4 - 6)^2$
$PQ^2 = (-1)^2 + (1)^2 + (-2)^2$
$PQ^2 = 1 + 1 + 4 = 6$

Answer: $6$