Equivalently, let the plane be given by Ax+ By+ Cz= D and the point be \(\displaystyle (x_0, y_0, z_0)\).

The "distance from point P to plane q" is, by definition, the **shortest** distance from P to any point in the plane.

And that must be along a line perpendicular to the plane- to see that choose any other point in the plane and draw the lines from P to those two points.

One, the perpendicular, is a leg of a right triangle while the other is a hypotenuse- and the hypotenuse of a right triangle is always longer than a leg.

A vector perpendicular to Ax+ By+ Cz= D is of the form <A, B, C> so a line through \(\displaystyle (x_0, y_0, z_0)\) in the direction of that line is \(\displaystyle x= At+ x_0\), \(\displaystyle y= Bt+ y_0\), and \(\displaystyle z= Ct+ z_0\). That line crosses the plane when A(At+ x_0)+ B(Bt+ y_0)+ C(Ct+ z_0)= D or \(\displaystyle (A^2+ B^2+ C^2)t+ (Ax_0+ By_0+ Cz_0)= D\)

so \(\displaystyle t= \frac{D- Ax_0- By_0- Cz_0}{A^2+ B^2+ C^2}\).

Put that value of t into the equation of the plane to find the point where the line crosses the plane and then find the distance between that point and \(\displaystyle (x_0, y_0, z_0)\).