## Problem Statement

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

$A(-7 ; 0 ; 3)$

$B(1 ;-5 ;-4)$

$C(2 ;-3 ; 0)$
## Solution

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

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

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:

$$
\begin{aligned}
& (x-(-7))+2 \cdot(y-0)+4 \cdot(z-3)=0 \\
& x+7+2 y+4 z-12=0 \\
& x+2 y+4 z-5=0
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 9-10