[ Coordinate method in space ] [ Parametric equations of a line ] Through the point $M(-2 ; 0 ; 3)$, draw a line that intersects the lines

$$
\left\{\begin{array} { l } 
{ x = 2 - t } \\
{ y = 3 } \\
{ z = - 2 + t }
\end{array} \left\{\begin{array}{l}
2 x-2 y-z-4=0 \\
x+3 y+2 z+1=0
\end{array}\right.\right.
$$
Through the point $M(-2 ; 0 ; 3)$ and the first line, we will construct the plane $\alpha$. For this, we will take any point on this line, for example, $A(2 ; 3 ;-2)$, find the vector $\overrightarrow{n_{1}}=(a ; b ; c)$, perpendicular to the direction vector $\overrightarrow{m_{1}}=(-1; 0 ; 1)$ and the vector $\overrightarrow{M A}=(4 ; 3 ;-5)$, and solve the system of equations

$$
\left\{\begin{array}{l}
\overrightarrow{n_{1}} \cdot \overrightarrow{m_{1}}=0 \\
\overrightarrow{n_{1}} \cdot \overrightarrow{M A}=0
\end{array}\right.
$$

or

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

Let $a=c=3$. Then $b=\frac{1}{3}(-4 a+5 c)=1$. Therefore, the equation of the desired plane is

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

We will find the parametric equations of the second given line. For this, we will set $x=t$ and solve the system of equations

$$
\left\{\begin{array}{l}
2 y+z=2 t-4 \\
3 y+2 z=-t-1
\end{array}\right.
$$

Multiplying both sides of the first equation by -2 and adding the result to the second equation, we get $y=5 t-7$. Therefore,

$$
z=2 t-4-2 y=2 t-4-10 t+14=-8 t+10
$$

Thus, we can take the vector $\overrightarrow{m_{2}}=(1 ; 5 ;-8)$ as the direction vector of the second line. Through the point $M(-2 ; 0 ; 3)$ and the second line, we will construct the plane $\beta$. For this, we will take any point on the second line, for example, $B(0 ;-7 ; 10)$, find the vector $\overrightarrow{n_{2}}=(p ; q ; r)$, perpendicular to the direction vector $\overrightarrow{m_{2}}=(1 ; 5 ;-8)$ and the vector $\overrightarrow{M B}=(2 ;-7 ; 7)$, and solve the system of equations

$$
\left\{\begin{array}{l}
\overrightarrow{n_{2}} \cdot \overrightarrow{m_{2}}=0 \\
\overrightarrow{n_{3}} \cdot \overrightarrow{M B}=0
\end{array}\right.
$$

or

$$
\left\{\begin{array}{l}
p+5 q-8 r=0 \\
2 p-7 q+7 r=0
\end{array}\right.
$$

Multiplying both sides of the first equation by -2 and adding the result to the second equation, we get $-17 q+23 r=0$. Let $q=23$. Then $r=17, p=8 r-5 q=21$. Therefore, the equation of the plane $\beta$ is

$$
21(x+2)+23 y+17(z-3)=0, \text { or } 21 x+23 y+17 z-9=0
$$

The desired line is the intersection of the planes $\alpha$ and $\beta$:

$$
\left\{\begin{array}{l}
3 x+y+3 z-3=0 \\
21 x+23 y+17 z-9=0
\end{array}\right.
$$

We will find its parametric equations. For this, we will set $y=-3 t$ and solve the system of equations

$$
\left\{\begin{array}{l}
3 x+3 z=3 t+3 \\
21 x+17 y=69 t+9
\end{array}\right.
$$

We get $x=13 t-2, z=3-12 t$.

## Answer

$x=-2+13 t, y=-3 t, z=3-12 t$.