JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Three Dimensional Geometry

For the first line, $\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}=\lambda$ gives the point $(2+2\lambda ,-2\lambda ,7+16\lambda )$.

For the second, $\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}=k$ gives $(-3+4k,-2+3k,-2+k)$.

Equating $x$ and $y$-coordinates:


From the second, $\lambda =\frac{2-3k}{2}$. Substituting into the first: $4-3k=-3+4k\Rightarrow 7k=7\Rightarrow k=1$, so $\lambda =-\frac{1}{2}$.

Checking $z$: $7+16\left(-\frac{1}{2}\right)=-1$ and $-2+1=-1$ — consistent. So $P=(1,1,-1)$.

The third line passes through $A(-1,1,1)$ with direction $\vec{b}=(2,3,1)$. Then $\overrightarrow{AP}=(2,0,-2)$.

The perpendicular distance from $P$ to this line satisfies


$\overrightarrow{AP}\cdot \vec{b}=4+0-2=2$, $|\vec{b}{|}^2=4+9+1=14$, $|\overrightarrow{AP}{|}^2=4+0+4=8$.