JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Three Dimensional Geometry

Let $A=(0,1,2)$ be the given point on L. The given line can be parametrized as $(1+2\lambda ,2+3\lambda ,3+4\lambda )$. Let $B$ be the intersection point, so $B=(1+2\lambda ,2+3\lambda ,3+4\lambda )$. The direction vector of L is $\overrightarrow{AB}=(1+2\lambda ,1+3\lambda ,1+4\lambda )$. For L to be parallel to the plane $2x+y-3z=4$, we need $\overrightarrow{AB}\cdot \vec{n}=0$ where $\vec{n}=(2,1,-3)$. Thus $2(1+2\lambda )+(1+3\lambda )-3(1+4\lambda )=2+4\lambda +1+3\lambda -3-12\lambda =-5\lambda =0$, so $\lambda =0$. Hence $B=(1,2,3)$ and $\overrightarrow{AB}=(1,1,1)$. The line L has equation $\vec{r}=(0,1,2)+\mu (1,1,1)$, i.e., a general point is $Q=(\mu ,1+\mu ,2+\mu )$. For the foot of perpendicular from $P(1,-9,2)$ to L, we need $\overrightarrow{PQ}\cdot \overrightarrow{AB}=0$: $(\mu -1,10+\mu ,\mu )\cdot (1,1,1)=(\mu -1)+(10+\mu )+\mu =3\mu +9=0$, so $\mu =-3$. The foot is $Q=(-3,-2,-1)$. The distance is $PQ=\sqrt{(1-(-3){)}^2+(-9-(-2){)}^2+(2-(-1){)}^2}=\sqrt{16+49+9}=\sqrt{74}$.