Distance from A (where a line and plane intersect) to P (which is a point perpendicular to the plane...

A coordinates = [-2, 1, 3]

P coordinates = [2, 3, -1]

(original equation of plane is 2x + y -2z + 9 = 0)

..thanks!

The way you have stated this, you are just asking for the distance from A to B: \(\displaystyle \sqrt{(-2-2)^2+ (1- 3)^2+ (3+ 1)^2}\).

That will be the "distance from point to plane" only if the line connecting the two points is perpendicular to the plane. The line from (-2, 1, 3) to (2, 3, -1) is given by x= -2+ 4t, y= 1+ 2t, z= 3- 4t which has "direction vector" <4, 2, -4>. That is, in fact, in the same direction as the normal vector, <2, 1, -2> of the plane so, apparently, you had already done the hard work!

The distance from point (2, 3, -1) to the plane 2x + y -2z + 9 = 0 is \(\displaystyle \sqrt{(-2-2)^2+ (1- 3)^2+ (3+ 1)^2}= \sqrt{16+ 4+ 16}= 6\)