## Problem Statement

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

$A(5, -1, 2)$

$B(2, -4, 3)$

$C(4, -1, 3)$
## Solution

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

$\overrightarrow{BC}=\{4-2 ;-1-(-4) ; 3-3\}=\{2 ; 3 ; 0\}$

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 of the form:
$2 \cdot(x-5)+3 \cdot(y+1)+0 \cdot(z-2)=0$

Simplify:

$$
\begin{aligned}
& 2 x-10+3 y+3=0 \\
& 2 x+3 y-7=0
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 9-12