## Problem Statement

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

$A(-7 ; 1 ;-4)$

$B(8 ; 11 ;-3)$

$C(9 ; 9 ;-1)$
## Solution

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

$$
\overrightarrow{B C}=\{9-8 ; 9-11 ;-1-(-3)\}=\{1 ;-2 ; 2\}
$$

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:
$(x-(-7))-2 \cdot(y-1)+2 \cdot(z-(-4))=0$

$x+7-2 y+2+2 z+8=0$

$x-2 y+2 z+17=0$

## Problem Kuznetsov Analytic Geometry 9-19