Let P be the plane, passing through the point $\left(1,-1,-5\right)$ and perpendicular to the line joining the points $\left(4,1,-3\right)$ and $\left(2,4,3\right)$.  Then the distance of P from the point $\left(3,-2,2\right)$ is

dr’s of  normal to the plane are $(2,-3,-6)=(\mathrm{a},\mathrm{b},\mathrm{c}),$

Point on the plane $\left({\mathrm{x}}_1,{\mathrm{y}}_1,{\mathrm{z}}_1\right)=(1,-1,-5)$
Equation of plane 'P'  is $\mathrm{a}\left(\mathrm{x}-{\mathrm{x}}_1\right)+\mathrm{b}\left(\mathrm{y}-{\mathrm{y}}_1\right)+\mathrm{c}\left(\mathrm{z}-{\mathrm{z}}_1\right)=0$
$2(x-1)-3(y+1)-6(z+5)=0\Rightarrow 2x-3y-6z=35$
distance from (3, -2,2) to plane P is $\left|\frac{6+6-12-35}{\sqrt{4+9+36}}\right|=\frac{35}{7}=5$