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 :

Let the required foot of the “oblique” perpendicular be the point $$Q$$ on the first line

$$\frac{x-1}{1}=\frac{y-1}{2}=\frac{z}{-1}\;,$$ whose direction vector is $$\mathbf{a}=(1,2,-1).$$ Parameterise this line as $$Q(1+t,\;1+2t,\;-t),\qquad t\in\mathbb{R}.$$

We must reach this line from the given point $$P(-2,-8,6)$$ along the second line

$$\frac{x+5}{1}=\frac{y+5}{-1}=\frac{z}{2},$$ whose direction vector is $$\mathbf{b}=(1,-1,2).$$ Hence the vector $$\overrightarrow{QP}=\mathbf{PQ}$$ has to be a scalar multiple of $$\mathbf{b}$$:

$$\overrightarrow{PQ}=P-Q=\lambda\mathbf{b},\qquad\lambda\in\mathbb{R}.$$

Write $$\overrightarrow{PQ}$$ component-wise:

$$P-Q=(-2-(1+t),\,-8-(1+2t),\,6-(-t))=(-3-t,\,-9-2t,\,6+t).$$

Equate each component to $$\lambda(1,-1,2)$$:

$$\begin{aligned} -3-t &= \lambda\quad &-(1)\\ -9-2t &= -\lambda\quad &-(2)\\ 6+t &= 2\lambda\quad &-(3) \end{aligned}$$

From $$(1)$$, $$\lambda=-3-t.$$ Insert this in $$(2)$$:

$$-9-2t = -(-3-t)=3+t\;\;\Longrightarrow\;\;-12-3t=0\;\;\Longrightarrow\;\;t=-4.$$

Therefore $$\lambda=-3-(-4)=1.$$ Check with $$(3)$$: $$6+t=6-4=2=2\lambda,\;$$ so all three equations agree.

Thus $$\overrightarrow{PQ}=1\cdot\mathbf{b}=\mathbf{b},$$ so the distance is

$$|PQ| = |\mathbf{b}| = \sqrt{1^{2}+(-1)^{2}+2^{2}}=\sqrt{6}.$$

The question asks for the square of this distance:

$$|PQ|^{2} = (\sqrt{6})^{2}=6.$$

Option B which is: 6