## Problem Statement

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

$A(-3 ; 6 ; 4)$

$B(8 ;-3 ; 5)$

$C(10 ;-3 ; 7)$
## Solution

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

$\overrightarrow{B C}=\{10-8 ;-3-(-3) ; 7-5\}=\{2 ; 0 ; 2\}$

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:

$$
\begin{aligned}
& 2 \cdot(x-(-3))+0 \cdot(y-6)+2 \cdot(z-4)=0 \\
& 2 x+6+2 z-8=0 \\
& 2 x+2 z-2=0 \\
& x+z-1=0
\end{aligned}
$$

## Problem Kuznetsov Analytic Geometry 8-30(2)

This problem may have a different condition (possibly due to different editions or errors). See above for more details.