## Problem Statement

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

$A(-3 ; 1 ; 0)$

$B(6 ; 3 ; 3)$

$C(9 ; 4 ;-2)$
## Solution

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

$\overrightarrow{BC}=\{9-6 ; 4-3 ;-2-3\}=\{3 ; 1 ;-5\}$

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:
$3 \cdot(x-(-3))+(y-1)-5 \cdot(z-0)=0$

$3 x+9+y-1-5 z=0$

$3 x+y-5 z+8=0$

## Problem Kuznetsov Analytic Geometry 9-21