JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Three Dimensional Geometry

For coplanarity, the direction ratios of $l_1$ (which are $3,1,-2$) must be perpendicular to the normal of the plane containing $l_2$. The plane through $l_2$ is $(3x+2y+z-2)+\mu (x-3y+2z-13)=0$. Its normal is $(3+\mu ,2-3\mu ,1+2\mu )$.

For perpendicularity: $3(3+\mu )+1(2-3\mu )-2(1+2\mu )=0$ gives $\mu =9/4$.

Substituting into the plane equation and using the point $(-5,-4,\alpha )$ on $l_1$, we find $\alpha =7$.

General point on $l_1$: $P(3\lambda -5,\lambda -4,-2\lambda +7)$. For $P$ nearest to $Q(-4,-3,2)$, $\overrightarrow{PQ}\perp l_1$:
$3(3\lambda -1)+1(\lambda -1)-2(-2\lambda +5)=0$ gives $\lambda =1$.

So $P(-2,-3,5)$ and $|a|+|b|+|c|=2+3+5=10$.