## Problem Statement

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

$A(2 ; 1 ; 7)$

$B(9 ; 0 ; 2)$

$C(9 ; 2 ; 3)$
## Solution

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

$\overrightarrow{BC}=\{9-9 ; 2-0 ; 3-2\}=\{0 ; 2 ; 1\}$

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:

$0 \cdot(x-2)+2 \cdot(y-1)+(z-7)=0$

$2 y-2+z-7=0$

$2 y+z-9=0$

## Problem Kuznetsov Analytic Geometry 9-18