## Problem Statement

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

$A(-3 ; 6 ; 4)$

$B(8 ;-3 ; 5)$

$C(0 ;-3 ; 7)$
## Solution

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

$\overrightarrow{BC}=\{0-8 ;-3-(-3) ; 7-5\}=\{-8 ; 0 ; 2\}$

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:

$-8 \cdot(x-(-3))+0 \cdot(y-6)+2 \cdot(z-4)=0$

$-8 x-24+2 z-8=0$

$-8 x+2 z-32=0$

$-4 x+z-16=0$