8,9

Find the distance between the line passing through points $A(-3 ; 0 ; 1)$ and $B(2 ; 1 ;-1)$, and the line passing through points $C(-2 ; 2 ; 0)$ and $D(1 ; 3 ; 2)$.
Let's find the coordinates of vectors $\overrightarrow{A B}$ and $\overrightarrow{C D}$:

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

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

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

Adding these equations term by term, we get the equation $8 a+2 b=0$, or $4 a+b=0$. Let $a=2$. Then $b=$ $-4 a=-8, c=\frac{1}{2}(5 a+b)=1$. Through the point $A(-3 ; 0 ; 1)$, we draw a plane perpendicular to the vector $\vec{n}=(2 ;-8 ; 1)$

$$
2(x+3)-8 y+z-1=0, \text { or } 2 x-8 y+z+5=0
$$

The distance from the point $C(-2 ; 2 ; 0)$ to this plane is equal to the distance between the lines $A B$ and $C D$. If $\rho$ is the desired distance, then

$$
\rho=\frac{|2 \cdot(-2)-8 \cdot 2+0+5|}{\sqrt{2^{2}+8^{2}+1^{2}}}=\frac{15}{\sqrt{69}}=\frac{5 \sqrt{3}}{\sqrt{23}}
$$

## Answer

$\sqrt{3}$

$\frac{\sqrt{3}}{\sqrt{23}}$