Find the distance from the point $D(1 ; 3 ; 2)$ to 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 $A$, draw a plane perpendicular to the vector $\vec{n}=(1 ; 1 ; 3):$

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

Let $\rho$ be the distance from point $D(x 0 ; y 0 ; z 0)$ to the plane given by the equation $a x+b y+c z+d=0$. Then

In our case

$$
\rho=\frac{|1+3+6|}{\sqrt{1^{2}+1^{2}+3^{2}}}=\frac{10}{\sqrt{11}}
$$

## Answer

$\frac{10}{\sqrt{11}}$.

## Problem