## Problem Statement

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

$A(4, -2, 0)$

$B(1, -1, -5)$

$C(-2, 1, -3)$
## Solution

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

$\overrightarrow{BC}=\{-2-1 ; 1-(-1) ;-3-(-5)\}=\{-3 ; 2 ; 2\}$

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:

$$
\begin{aligned}
& -3 \cdot(x-4)+2 \cdot(y+2)+2 \cdot(z-0)=0 \\
& -3 x+12+2 y+4+2 z=0 \\
& -3 x+2 y+2 z+16=0
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 9-3