Question

If the distance between the plane, $23x-10y-2z+48=0$ and the plane containing the lines $\frac{x+1}{2}=\frac{y-3}{4}=\frac{z+1}{3}$ and $\frac{x+3}{2}=\frac{y+2}{6}=\frac{z-1}{\lambda },(\lambda \in R)$ is equal to $\frac{k}{\sqrt{633}},$ then $k$ is equal to

Answer 1

$\because$ Both the lines lie in the plane which is parallel to the plane $23x-10y-2z+48=0$
$\Rightarrow$ both the line will be parallel to the given plane
$\Rightarrow$ distance from any point on the line to given plane will be $=\frac{k}{\sqrt{633}}$
$\Rightarrow$ distance of the given plane from the point $(-1,3,-1)=\frac{k}{\sqrt{633}}\Rightarrow \frac{|(-23-30+2+48)|}{\sqrt{{23}^2+{10}^2+2^2}}=\frac{k}{\sqrt{633}}\Rightarrow \frac{3}{\sqrt{633}}=\frac{k}{\sqrt{633}}\Rightarrow k=3$