If a line having direction ratios  intersect lines $\frac{\mathrm{x}-7}{3}=\frac{\mathrm{y}-1}{-1}=\frac{\mathrm{z}+2}{1},\frac{\mathrm{x}}{2}=\frac{\mathrm{y}-7}{3}=\frac{\mathrm{z}}{1}$  at point A, B respectively. Then the value of (AB)² is 

The given lines are $L_1\equiv \frac{\mathrm{x}-7}{3}=\frac{\mathrm{y}-1}{-1}=\frac{\mathrm{z}+2}{1}$ and $L_2\equiv \frac{x}{2}=\frac{y-7}{3}=\frac{z}{1}$

Let $A,B$ are general points on the above two lines

Hence, $A\left(3k+7,-k+1,k-2\right),B\left(2s,3s+7,s\right)$

Direction ratios of $\stackrel{\rightharpoonup }{AB}$ are 

These are proportional with the given direction ratios 

It implies that $\frac{2s-3k-7}{1}=\frac{3s+k+6}{-4}=\frac{s-k+2}{2}$

Form first two equtaions

We get

             $\begin{array}{rcl}-8s+12k+28& =& 3s+k+6\\ 11s-11k-22& =& 0\\ s-k& =& 2\end{array}$

From last two equations 

we get 

          $\begin{array}{rcl}6s+2k+12& =& -4s+4k-8\\ 10s-2k& =& -20\\ 5s-k& =& -10\end{array}$

Subtracting one equation from the other equation

we get 

$\begin{array}{rcl}5s-k-s+k& =& -10-2\\ 4s& =& -12\\ s& =& -3\end{array}$

Substitute $s=-3$ in the equation $s-k=12$

Hence $k=-5$

The points are $A\left(-15+7,5+1,-5-2\right)=\left(-8,6,-7\right)$ and $B\left(-6,-2,-3\right)$

${\left(AB\right)}^2={\left(2\right)}^2+{\left(8\right)}^2+{\left(4\right)}^2=\boxed{84}$