Thank you for reporting, we will resolve it shortly
Q.
Let $P$ be the plane, passing through the point $(1,-1,-5)$ and perpendicular to the line joining the points $(4,1,-3)$ and $(2,4,3)$. Then the distance of $P$ from the point $(3,-2,2)$ is
Equation of Plane :
$2( x -1)-3( y +1)-6( z +5)=0$
Or $2 x-3 y-6 z=35$
$\Rightarrow$ Required distance $=$
$\frac{|2(3)-3(-2)-6(2)-35|}{\sqrt{4+9+36}}$
$=5$