## Task Condition

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

$A(0; -2; 8)$

$B(4; 3; 2)$

$C(1; 4; 3)$
## Solution

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

$\overrightarrow{B C}=\{1-4 ; 4-3 ; 3-2\}=\{-3 ; 1 ; 1\}$

Since the vector $\overrightarrow{B C}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:

$$
\begin{aligned}
& -3 \cdot(x-0)+(y-(-2))+(z-8)=0 \\
& -3 x+y+2+z-8=0 \\
& -3 x+y+z-6=0
\end{aligned}
$$

Kuznetsov Analytical Geometry 9-14