## Problem Statement

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

$A(2 ; 5 ; -3)$

$B(7 ; 8 ; -1)$

$C(9 ; 7 ; 4)$
## Solution

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

$\overrightarrow{B C}=\{9-7 ; 7-8 ; 4-(-1)\}=\{2 ;-1 ; 5\}$

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:

$$
2 \cdot(x-2)-(y-5)+5 \cdot(z-(-3))=0
$$

Simplify:

$$
\begin{aligned}
& 2 x-4-y+5+5 z+15=0 \\
& 2 x-y+5 z+16=0
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 9-31