## Problem Statement

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

$A(7, -5, 1)$

$B(5, -1, -3)$

$C(3, 0, -4)$
## Solution

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

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

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:

$$
\begin{aligned}
& -2 \cdot(x-7)+(y-(-5))-(z-1)=0 \\
& -2 x+14+y+5-z+1=0 \\
& -2 x+y-z+20=0
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 9-5