Let $A=(-2,-8,6)$. The line through $A$ with direction $(1,-1,2)$ (parallel to the second given line) is

$\frac{x+2}{1}=\frac{y+8}{-1}=\frac{z-6}{2}=\lambda .$

Its general point is $A+\lambda (1,-1,2)=(\lambda -2,-\lambda -8,2\lambda +6)$.

This must meet the first line, parametrised as $(\mu +1,2\mu +1,-\mu )$. Equating coordinates:

$\lambda -2=\mu +1,-\lambda -8=2\mu +1,2\lambda +6=-\mu .$

From the first, $\mu =\lambda -3$. Substitute into the third: $2\lambda +6=-(\lambda -3)=3-\lambda \Rightarrow 3\lambda =-3\Rightarrow \lambda =-1$, $\mu =-4$ (which also satisfies the second).

Intersection point $B=(-3,-7,4)$ and

$A{B}^2=(-3+2{)}^2+(-7+8{)}^2+(4-6{)}^2=1+1+4=6.$