Example 1. Reduce the general equations of a line to canonical form

\[
\left\{\begin{array}{l}
2 x-3 y-3 z-9=0 \\
x-2 y+z+3=0
\end{array}\right.
\]
Solution. We will form the canonical equations of the line using two points (as in point $4^{\circ}$). The coordinates of the two points on the line will be found according to the scheme in point $5^{\circ}$.

1) Let, for example, $z_{1}=0$ and solve the system

$$
\left\{\begin{array}{l}
2 x-3 y-9=0, \\
x-2 y+3=0
\end{array} \quad \Rightarrow x_{1}=27, y_{1}=15\right.
$$

The point $M_{1}(27,15,0)$ lies on the line.

2) Similarly, let $z_{2}=-3$. Then

$$
\left\{\begin{array}{l}
2 x-3 y=0, \\
x-2 y=0
\end{array} \Rightarrow x_{2}=0, y_{2}=0\right.
$$

The point $M_{2}(0,0,-3)$ also belongs to the line.

3) Write the equations of the line passing through the two points (point $4^{\circ}$):

$$
\frac{x}{27}=\frac{y}{15}=\frac{z+3}{3}, \quad \text { or } \quad \frac{x}{9}=\frac{y}{5}=\frac{z+3}{1}
$$