$L$ is a line passing through the point $A(1,0,-3)$ and parallel to a line having direction ratios $0,1,-2…
$L$ is a line passing through the point $A(1,0,-3)$ and parallel to a line having direction ratios $0,1,-2 P$ is a point on the line $L$ which is at a minimum distance from the plane $2 x+3 y+5 z=1$. Then, the equation of the plane through $P$ and perpendicular to $A P$ is
$y+2 z=12$
$y-2 z+4=0$
$x+y-2 z=12$
$2 y-z=16$
Solution
Equation of line $L$ is $\frac{x-1}{0}=\frac{y-0}{1}=\frac{z+3}{-2}=\lambda$ (say)
Then, any point on $L$ is of the form $(1, \lambda,-2 \lambda-3)$. Now, the distance of $(1, \lambda,-2 \lambda-3)$ from the plane $2 x+3 y+5 z=1$ is
$\begin{aligned}
d & =\frac{|2(1)+3 \lambda+5(-2 \lambda-3)-1|}{\sqrt{2^2+3^2+5^2}} \\
& =\frac{|2+3 \lambda-10 \lambda-15-1|}{\sqrt{38}} \\
& =\frac{|-7 \lambda-14|}{\sqrt{38}}=\frac{|7 \lambda+14|}{\sqrt{38}}
\end{aligned}$
Clearly, $d$ will be minimum when $7 \lambda+14=0$ i.e. $\lambda=-2$. Thus, the coordinates of $P$ are $(1,-21)$ and DR's of AP are $1-1,-2-0,1+3$ i.e. $0,-2,4$ Now, equation of plane through $P$ and perpendicular to $\mathrm{AP}$ is
$\begin{aligned}
\Rightarrow & & 0 \cdot(x-1)-2(y+2)+4(z-1) & =0 \\
\Rightarrow & & -2 y-4+4 z-4 & =0 \\
\Rightarrow & & 2 y-4 z=-8 \Rightarrow y-2 z+4 & =0
\end{aligned}$