## Task Condition

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

$A(0; -3; 5)$

$B(-7; 0; 6)$

$C(-3; 2; 4)$
## Solution

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

$\overrightarrow{BC}=\{-3-(-7) ; 2-2 ; 4-6\}=\{4 ; 0 ;-2\}$

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:
$4 \cdot(x-0)+0 \cdot(y-(-3))-2 \cdot(z-5)=0$

$4 x-2 z+10=0$

$2 x-z+5=0$

## Problem Kuznetsov Analytic Geometry 9-11