If P(2, 3, 1) is a point and $L\equiv x-y-z-2=0$ is a plane then

At (0,0,0), x – y – z – 2 = –2
= (–ve) at (2,3,1), x – y – z – 2 = 2 –3 – 1 – 2 = – 4
Since, both have same sign (0, 0, 0) and
(2, 3, 1) lie on the same side of the plane.
Distance $=\left|\frac{2-3-1-2}{\sqrt{1^2+1^2+1^2}}\right|=\frac{4}{\sqrt{3}}$
Equation of a line perpendicular to the plane
x –y –z –2 = 0 and passing through the point $(2,3,1)\text{ is }\frac{x-2}{1}\equiv \frac{y-3}{-1}\equiv \frac{z-1}{-1}=\lambda$
A point on the line is $(\lambda +2,3-\lambda ,1-\lambda )$ and it lies
and the plane x – y – z – 2 = 0
if  $\lambda +2-3+\lambda -1=\lambda -2=0\Rightarrow \frac{4}{3}$
Foot of perpendicular on the plane is
$\left(\frac{4}{3}+2,3-\frac{4}{3},1-\frac{4}{3}\right)=\left(\frac{10}{3},\frac{5}{3},-\frac{1}{3}\right)$