## Problem Statement

Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.

$A(-10 ; 0 ; 9)$

$B(12 ; 4 ; 11)$

$C(8 ; 5 ; 15)$
## Solution

Let's find the vector $\overrightarrow{BC}$:

$\overrightarrow{BC}=\{8-12 ; 5-4 ; 15-11\}=\{-4 ; 1 ; 4\}$

Since the vector $\overrightarrow{BC}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:

$-4 \cdot(x+10) + (y-0) + 4 \cdot(z-9) = 0$

Simplify:

$-4x - 40 + y + 4z - 36 = 0$

$-4x + y + 4z - 76 = 0$

Kuznetsov Analytical Geometry 9-16