Find the distance of the point (2, 3, -5) from the plane x + 2y - 2z - 9 = 0 ?

Concept:

Perpendicular Distance of a Point from a Plane 

Let us consider a plane given by the Cartesian equation, Ax + By + Cz = d and a point whose coordinate is, (x1, y1, z1). Then the distance between the point and the plane is given by 

Calculation:

We have to find the distance of the point (2, 3, -5) from the plane x + 2y - 2z = 9 

As we know that,  the distance between the point and the plane is given by 

Here, A = 1, B = 2, C = -2, d = 9, x1 = 2, y1 = 3 and z1 = - 5

So, the distance of the given point form the given plane 

So, the distance between the plane and the point is 3 units

Hence, option D is the correct answer.