## Problem Statement

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

$A(-1 ; 3 ; 4)$

$B(-1 ; 5 ; 0)$

$C(2 ; 6 ; 1)$
## Solution

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

$\overrightarrow{BC}=\{2-(-1) ; 6-5 ; 1-0\}=\{3 ; 1 ; 1\}$

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:
$3 \cdot(x-(-1))+(y-3)+(z-4)=0$

$3 x+3+y-3+z-4=0$

$3 x+y+z-4=0$

Kuznetsov Analytical Geometry 9-2