## problem statement

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

$A(1 ; 0 ;-6)$

$B(-7 ; 2 ; 1)$

$C(-9 ; 6 ; 1)$
## Solution

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

$\overrightarrow{B C}=\{-9-(-7) ; 6-2 ; 1-1\}=\{-2 ; 4 ; 0\}$

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:

$$
\begin{aligned}
& -2 \cdot(x-1)+4 \cdot(y-0)+0 \cdot(z-(-6))=0 \\
& -2 x+2+4 y=0 \\
& -x+2 y+1=0
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 9-20