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

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 $\lambda=-5, \mu=-3$
$A =(-8,6,7)$
$B =(-6,-2,-3)$
$( AB )^{2}=4+64+16=84$