## Task Condition

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

$A(1; -5; -2)$

$B(6; -2; 1)$

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

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

$\overrightarrow{BC}=\{2-6 ;-2-(-2) ;-2-1\}=\{-4 ; 0 ;-3\}$

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:

$-4 \cdot(x-1)+0 \cdot(y-(-5))-3 \cdot(z-(-2))=0$

$-4 x+4+-3 z-6=0$

$-4 x-3 z-2=0$

$4 x+3 z+2=0$

Kuznetsov Analytical Geometry 9-24