## Problem Statement

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

$A(-8 ; 0 ; 7)$

$B(-3 ; 2 ; 4)$

$C(-1 ; 4 ; 5)$
## Solution

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

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

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:

$$
\begin{aligned}
& 2 \cdot(x-(-8))+2 \cdot(y-0)+(z-7)=0 \\
& 2 x+16+2 y+z-7=0 \\
& 2 x+2 y+z+9=0
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 9-4