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

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

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

The equation of the line passing through $P$P and parallel to this line is:

$$\dfrac{x+2}{1} = \dfrac{y+8}{-1} = \dfrac{z-6}{2} = \lambda$$1x+2​=−1y+8​=2z−6​=λ

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

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

$$\dfrac{\lambda - 2 - 1}{1} = \dfrac{-\lambda - 8 - 1}{2} = \dfrac{2\lambda + 6}{-1}$$1λ−2−1​=2−λ−8−1​=−12λ+6​

$$\dfrac{\lambda - 3}{1} = \dfrac{-\lambda - 9}{2} = \dfrac{2\lambda + 6}{-1}$$1λ−3​=2−λ−9​=−12λ+6​

Taking the first two parts:

$$2(\lambda - 3) = -\lambda - 9$$2(λ−3)=−λ−9

$$2\lambda - 6 = -\lambda - 9$$2λ−6=−λ−9

$$3\lambda = -3 \Rightarrow \lambda = -1$$3λ=−3⇒λ=−1

Substituting $\lambda = -1$λ=−1 into the coordinates of $Q$Q, we get:

$$Q(-1 - 2, -(-1) - 8, 2(-1) + 6) \equiv Q(-3, -7, 4)$$Q(−1−2,−(−1)−8,2(−1)+6)≡Q(−3,−7,4)

The square of the distance between $P(-2, -8, 6)$P(−2,−8,6) and $Q(-3, -7, 4)$Q(−3,−7,4) is:

$$PQ^2 = (-3 - (-2))^2 + (-7 - (-8))^2 + (4 - 6)^2$$PQ2=(−3−(−2))2+(−7−(−8))2+(4−6)2

$$PQ^2 = (-1)^2 + (1)^2 + (-2)^2$$PQ2=(−1)2+(1)2+(−2)2

$$PQ^2 = 1 + 1 + 4 = 6$$PQ2=1+1+4=6

Answer: $6$6