We have to find the distance of the point (2, 3, -5) from the plane x + 2y - 2z - 9 = 0 As we know that, the distance between the point and the plane is given by |
Ax1+By1+Cz1−d
√A2+B2+C2
| Here, x1=2,y1=3 and z1=−5 So, the distance of the given point form the given plane = |
2×1+2×3+2×5−9
√12+(−2)2+(2)2
| =|
9
√9
| So, the distance between the plane and the point is 3 units Hence, option D is the correct answer.