## Problem Statement

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

$A(-1 ; 2 ; -2)$

$B(13 ; 14 ; 1)$

$C(14 ; 15 ; 2)$
## Solution

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

$\overrightarrow{B C}=\{14-13 ; 15-14 ; 2-1\}=\{1 ; 1 ; 1\}$

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:

$$
\begin{aligned}
& (x-(-1))+(y-2)+(z-(-2))=0 \\
& x+1+y-2+z+2=0 \\
& x+y+z+1=0
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 9-28