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

Given,

$P$ 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 $\left(2,4,3\right)$,

So, direction ratio of Plane Normal will be, $\left(4-2,1-4,-3-3\right)\equiv \left(2,-3,-6\right)$

Now, equation of Plane will be,

$2\left(x-1\right)-3\left(y+1\right)-6\left(z+5\right)=0$

$\Rightarrow 2x-3y-6z-35=0$

Hence, perpendicular distance of plane from $\left(3,-2,2\right)$ will be,

Distance $=\left|\frac{2\times 3-3\times \left(-2\right)-6\times 2-35}{\sqrt{2^2+{\left(-3\right)}^2+{\left(-6\right)}^2}}\right|=\left|\frac{6+6-12-35}{\sqrt{2^2+{\left(-3\right)}^2+{\left(-6\right)}^2}}\right|=5$.