## Task Condition

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

$A(0 ; 7 ;-9)$

$B(-1 ; 8 ;-11)$

$C(-4 ; 3 ;-12)$
## Solution

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

$\overrightarrow{B C}=\{-4-(-1) ; 3-8 ;-12-(-11)\}=\{-3 ;-5 ;-1\}$

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:

$$
\begin{aligned}
& -3 \cdot(x-0)-5 \cdot(y-7)-(z-(-9))=0 \\
& -3 x-5 y+35-z-9=0 \\
& -3 x-5 y-z+26=0
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 9-25