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

Step-by-step solution

1. Determine the normal vector of the plane $\vec{n}=(2,,3,,-1)$.
2. Use the distance formula $d=\frac{|A{x}_0+B{y}_0+C{z}_0+D|}{\sqrt{A^2+B^2+C^2}}$ where $D=-10$.
3. Substitute $x_0=3,,y_0=-2,,z_0=5$ to get $d=\frac{|2\cdot 3+3\cdot (-2)-1\cdot 5-10|}{\sqrt{2^2+3^2+(-1{)}^2}}$.
4. Simplify the numerator $|6-6-5-10|=15$ and the denominator $\sqrt{4+9+1}=\sqrt{14}$.
5. Thus $d=\frac{15}{\sqrt{14}}$.

Concept tested: Distance from point to plane

Published 13 September 2026 · How this solution was prepared and checked