## Problem Statement

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

$A(-4, -2, 5)$

$B(3, -3, -7)$

$C(9, 3, -7)$
## Solution

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

$\overrightarrow{BC}=\{9-3 ; 3-(-3) ;-7-(-7)\}=\{6 ; 6 ; 0\}$

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:

$6 \cdot(x-(-4))+6 \cdot(y-(-2))+0 \cdot(z-5)=0$

$6 x+24+6 y+12=0$

$x+y+6=0$

## Problem Kuznetsov Analytic Geometry 9-22