$\left[\begin{array}{l}\text { Coordinate method in space }] \\ {[\quad \underline{\text { equation of a plane }}]}\end{array}\right]$

Form the equation of the plane passing through the points $A(-3 ; 0 ; 1), B(2 ; 1 ;-1)$ and $C(-2 ; 2 ; 0)$.
Let's find the coordinates of vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:

$$
\begin{aligned}
& \overrightarrow{A B}=(2-(-3) ; 1-0 ;-1-1)=(5 ; 1 ;-2) \\
& \overrightarrow{A C}=(-2-(-3) ; 2-0 ; 0-1)=(1 ; 2 ;-1)
\end{aligned}
$$

Let $\vec{n}=(a ; b ; c)$ be a non-zero vector perpendicular to the desired plane. Then $\vec{n} \cdot \overrightarrow{A B}=0$ and $\vec{n} \cdot \overrightarrow{A C}=0$, or

$$
\left\{\begin{array}{l}
5 a+b-2 c=0 \\
a+2 b-c=0
\end{array}\right.
$$

Multiply both sides of the second equation by -2 and add the result term by term to the first. We get the equation $3 a-3 b=0$. Let $a=b=1$. Then $c=a+2 b=3$. Through point $A$, we draw a plane perpendicular to the vector $\vec{n}=(1 ; 1 ; 3):$

$$
x+3+y+3(z-1)=0, \text { or } x+y+3 z=0 \text { . }
$$

## Answer

$x+y+3 z=0$