## problem statement

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

$A(0; -8; 10)$

$B(-5; 5; 7)$

$C(-8; 0; 4)$
## Solution

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

$\overrightarrow{B C}=\{-8-(-5) ; 0-5 ; 4-7\}=\{-3 ;-5 ;-3\}$

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:

$$
\begin{aligned}
& -3 \cdot(x-0)-5 \cdot(y-(-8))-3 \cdot(z-10)=0 \\
& -3 x-5 y-40-3 z+30=0 \\
& -3 x-5 y-3 z-10=0 \\
& 3 x+5 y+3 z+10=0
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry $9-23$