## Task Condition

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

$A(1; -1; 8)$

$B(-4; -3; 10)$

$C(-1; -1; 7)$
## Solution

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

$\overrightarrow{BC}=\{-1-(-4) ;-1-(-3) ; 7-10\}=\{3 ; 2 ;-3\}$

Since the vector $\overrightarrow{BC}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:

$3 \cdot(x-1)+2 \cdot(y-(-1))-3 \cdot(z-8)=0$

$3 x-3+2 y+2-3 z+24=0$

$3 x+2 y-3 z+23=0$

## Problem Kuznetsov Analytic Geometry 9-7