## Problem Statement

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

$A(-3, -1, 7)$

$B(0, 2, -6)$

$C(2, 3, -5)$
## Solution

Let's find the vector $\overrightarrow{BC}:$
$\overrightarrow{BC}=\{2-0 ; 3-2 ;-5-(-6)\}=\{2 ; 1 ; 1\}$

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:

$2 \cdot(x-(-3))+(y-(-1))+(z-7)=0$

$2 x+6+1 y+1+z-7=0$

$2 x+y+z=0$

## Problem Kuznetsov Analytic Geometry 9-26