## Problem Statement

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

$A(3, -3, -6)$

$B(1, 9, -5)$

$C(6, 6, -4)$
## Solution

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

$\overrightarrow{BC}=\{6-1 ; 6-9 ;-4-(-5)\}=\{5 ;-3 ; 1\}$

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:

$5 \cdot(x-3)-3 \cdot(y-(-3))+(z-(-6))=0$

$5 x-15-3 y-9+z+6=0$

$5 x-3 y+z-18=0$

## Problem Kuznetsov Analytic Geometry 9-17