## Problem Statement

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

$A(-3 ; 7 ; 2)$

$B(3 ; 5 ; 1)$

$C(4 ; 5 ; 3)$
## Solution

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

$\overrightarrow{BC}=\{4-3 ; 5-5 ; 3-1\}=\{1 ; 0 ; 2\}$

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:

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

Simplify:

$x+3+2z-4=0$

$x+2z-1=0$

## Problem Kuznetsov Analytic Geometry 9-13