Question

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 point $A$ and $B$. Then $(AB{)}^2$ is equal to ____

Answer 1

Answer: 84

DR's of $AB$
$(3\lambda -2\mu +7,-\lambda -3\mu -6,\lambda -\mu -2)$
$\frac{3\lambda -2\mu +7}{1}=\frac{-\lambda -3\mu -6}{-4}=\frac{\lambda -\mu -2}{2}$
Taking first (2) $-12\lambda +8\mu -28=-\lambda -3\mu -6$
$\lambda -\mu +2=0$
Taking second & third
$-2\lambda -6\mu -12=-4\lambda +4\mu +8$
$\lambda -5\mu -10=0$
After solving above two equation $\lambda =-5,\mu =-3$
$A=(-8,6,7)$
$B=(-6,-2,-3)$
$(AB{)}^2=4+64+16=84$