## Problem Statement

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

$A(1 ; 0 ; -2)$

$B(2 ; -1 ; 3)$

$C(0 ; -3 ; 2)$
## Solution

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

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

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:

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

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

$-2x - 2y - z = 0$

$2x + 2y + z = 0$

## Problem Kuznetsov Analytic Geometry $9-1$