Calculate the perpendicular distance from the point $Q (4, - 2, 1)$ to the plane $3 x + 2 y - 5 z = 12$ .

medium 3 marks Class 12 · Mathematics · Three Dimensional Geometry

Answer

\frac{9}{38}

The correct answer is $\frac{9}{38}$ — Divide the numerator by this magnitude to obtain the distance: $\frac{9}{38}$ .

Step-by-step solution

Rewrite the plane equation in the form $A x + B y + C z + D = 0$ ; here $A = 3$ , $B = 2$ , $C = - 5$ , $D = - 12$ .
Substitute $x = 4$ , $y = - 2$ , $z = 1$ into $A x + B y + C z + D$ to get the numerator: $3 (4) + 2 (- 2) - 5 (1) - 12 = - 9$ .
Take the absolute value: $∣ - 9∣ = 9$ .
Compute the magnitude of the normal vector: $A^{2} + B^{2} + C^{2} = 9 + 4 + 25 = 38$ .
Divide the numerator by this magnitude to obtain the distance: $\frac{9}{38}$ .

Common mistake: Using the wrong sign for the plane constant

D

or forgetting the absolute value.

Concept tested: Distance from point to plane

Published 6 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked