Find the perpendicular distance of the point $P(3,4,5)$ from the plane $2x-3y+z=10$.

Step-by-step solution

1. Write the distance formula for point to plane: $d=\frac{|A{x}_0+B{y}_0+C{z}_0+D|}{\sqrt{A^2+B^2+C^2}}$.
2. Substitute $A=2,B=-3,C=1,D=-10$ and point $(x_0,y_0,z_0)=(3,4,5)$ into the numerator: $|2(3)-3(4)+1(5)-10|=|6-12+5-10|=|-11|=11$.
3. Substitute $A,B,C$ into the denominator: $\sqrt{2^2+(-3{)}^2+1^2}=\sqrt{4+9+1}=\sqrt{14}$.
4. Divide: $d=\frac{11}{\sqrt{14}}$.

Concept tested: Distance from point to plane

Published 17 September 2026 · How this solution was prepared and checked