JEE Main 2023 — Three Dimensional Geometry Question with Solution

Given,

The plane $P:8x+{\alpha }_{1}y+{\alpha }_{2}z+12=0$ be parallel to the line $L:\frac{x+2}{2}=\frac{y-3}{3}=\frac{z+4}{5}$ and the intercept of $P$ on the $y$-axis is $1$,

Now normal of the plane will be perpendicular to the given line, so by perpendicular condition we get,

$\Rightarrow 8\times 2+3{\alpha }_1+5{\alpha }_2=0 \dots \left(1\right)$

For $Y$ intercept of plane put $x=z=0$ in equation of plane

$\Rightarrow y=\frac{-12}{{\alpha }_1}=1\Rightarrow {\alpha }_1=-12$

Put ${\alpha }_1$ in equation $\left(1\right)$ we get ${\alpha }_2=4$

Hence, equation of plane $P: 2 x-3 y+z+3=0$

Now distance of point $A\left(-2,3,-4\right)$ from the plane $P$ is $=\left|\frac{2\left(-2\right)-3\left(3\right)-4+3}{\sqrt{{\left(2\right)}^2+{\left(-3\right)}^2+{\left(1\right)}^2}}\right|=\sqrt{14}$.