Find the perpendicular distance of the point $P(2,-3,6)$ from the plane $3x-4y+12z=30$.

Step-by-step solution

1. Identify the normal vector of the plane as $\mathbf{n}=⟨3,-4,12⟩$.
2. Use the point‑plane distance formula $d=\frac{|a{x}_0+b{y}_0+c{z}_0-d|}{\sqrt{a^2+b^2+c^2}}$ where $(x_0,y_0,z_0)$ is the point and $ax+by+cz=d$ is the plane.
3. Substitute $a=3$, $b=-4$, $c=12$, $d=30$, and $(x_0,y_0,z_0)=(2,-3,6)$:
4. Compute the numerator: $|3(2)-4(-3)+12(6)-30|=|6+12+72-30|=|60|=60$.
5. Compute the denominator: $\sqrt{3^2+(-4{)}^2+12^2}=\sqrt{9+16+144}=\sqrt{169}=13$.
6. Divide numerator by denominator: $d=\frac{60}{13}$.
7. State the distance with correct units.

Concept tested: Distance of a point from a plane

Published 14 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked