## Problem Statement

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

$A(-2 ; 0 ;-5)$

$B(2 ; 7 ;-3)$

$C(1 ; 10 ;-1)$
## Solution

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

$\overrightarrow{BC}=\{1-2 ; 10-7 ;-1-(-3)\}=\{-1 ; 3 ; 2\}$

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:

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

$-x-2+3 y+2 z+10=0$

$-x+3 y+2 z+8=0$

## Problem Kuznetsov Analytic Geometry 9-8