## Task Condition

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

$A(-3 ; 5 ;-2)$

$B(-4 ; 0 ; 3)$

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

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

$\overrightarrow{B C}=\{-3-(-4) ; 2-0 ; 5-3\}=\{1 ; 2 ; 2\}$

Since the vector $\overrightarrow{B C}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:

$$
\begin{aligned}
& (x-(-3))+2 \cdot(y-5)+2 \cdot(z-(-2))=0 \\
& x+3+2 y-10+2 z+4=0 \\
& x+2 y+2 z-3=0
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 9-6