Let a line having direction ratios $1,-4,2$ intersect the lines $\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1}$ and $\frac{x}{2}=\frac{y-7}{3}=\frac{z}{1}$ at the points $A$ and $B$. Then ${\left(AB\right)}^2$ is equal to

Let $A$ be $\left(3\lambda +7,-\lambda +1,\lambda -2\right)$ and $B$ be $\left(2\mu ,3\mu +7,\mu \right)$

DR's of $AB$ is

$\left(3\lambda -2\mu +7,-\lambda -3\mu -6,\lambda -\mu -2\right)$

Now, 

$\frac{3\lambda -2\mu +7}{1}=\frac{-\lambda -3\mu -6}{-4}=\frac{\lambda -\mu -2}{2}$

i.e. $-12\lambda +8\mu -28=-\lambda -3\mu -6$

$\Rightarrow \lambda =\mu -2$

Also 

$-2\lambda -6\mu -12=-4\lambda +4\mu +8$

$\Rightarrow \lambda =5\mu +10$

After solving above two equations, we get $\lambda =-5,\mu =-3$

i.e. $A=\left(-8,6,-7\right)$ and $B=\left(-6,-2,-3\right)$

Hence, ${\left(AB\right)}^2=4+64+16=84$