Consider two lines $L_1=\frac{x-7}{3}=\frac{y-7}{2}=\frac{z-3}{1}$ and $L_2=\frac{x-1}{2}=\frac{y+1}{4}=\frac{z+1}{3}$, $L$is a line  whose direction ratios are $\left(2,2,1\right)$intersect the lines $L_1$and $L_2$at A and B, then find the distance  $\frac{AB}{2}$

general point on $L_1=(3\lambda +7,2\lambda +7,\lambda +3)$

General point on  $L_2=(2\mu +1,4\mu -1,3\mu -1)$

DR of line joining the point A and B

$<3\lambda -2\mu +6,2\lambda -4\mu +8,\lambda -3\mu +4>$

DR of AB º<  2,2,1> (given)

$\frac{3\lambda -2\mu +6}{2}=\frac{2\lambda -4\mu +8}{2}=\frac{\lambda -3\mu +4}{1}$

$\text{On solving }\lambda =2,\mu =0$

$A(13,11,5)B(1,-1,-1)$

$AB=\sqrt{144+144+36}=\sqrt{324}=18$

So    $\frac{AB}{2}=\left(9\right)$