JEE Main 2023 (Online) 12th April Morning ShiftQuestion 1
Let the lines $l_1:\frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}$ and $l_2:3x+2y+z-2=0=x-3y+2z-13$ be coplanar. If the point $P(a,b,c)$ on $l_1$ is nearest to the point $Q(-4,-3,2)$, then $|a|+|b|+|c|$ is equal to
3D Geometry
Solution
Direction of $l_2$: $(3,2,1)\times(1,-3,2)=(7,-5,-11)$, point $(-\frac95,0,\frac{37}5)$. Coplanarity gives $\alpha=7$. Point $(-5+3s,-4+s,7-2s)$; $\overrightarrow{QP}\cdot(3,1,-2)=14s-14=0$, $s=1$, $P=(-2,-3,5)$. So $|a|+|b|+|c|=10$.