412. Find the direction cosines of the line

$$
\left\{\begin{array}{c}
2 x-3 y-3 z-9=0 \\
x-2 y+z+3=0
\end{array}\right.
$$
Solution. First, we convert the equations of the line to the canonical form. For this, in the given system of equations, we eliminate the variable $x$ and express $z$ in terms of $y$:

$$
\left\{\begin{array}{c|c}
2 x-3 y-3 z-9=0, & 1 \\
x-2 y+z+3=0 ; & -2
\end{array} \begin{array}{r}
2 x-3 y-3 z-9=0 \\
-2 x+4 y-2 z-6=0 \\
\hline-5 z-15=0
\end{array}\right.
$$

From this,

$$
z=\frac{y-15}{5}
$$

Now, express $z$ in terms of $x$:

$$
\left\{\begin{array}{c|r|r}
2 x-3 y-3 z-9=0, & 2 & 4 x-6 y-6 z-18=0 \\
x-2 y+z+3=0 ; & -3 & \frac{-3 x+6 y-3 z-9=0}{x-9 z-27=0}
\end{array}\right.
$$

From this,

$$
z=\frac{x-27}{9}
$$

Considering expressions (A) and (B), we write the equations of the line in canonical form:

$$
\frac{x-27}{9}=\frac{y-15}{5}=\frac{z}{1}
$$

It is easy to see that the line passes through the point $M(27 ; 15 ; 0)$, with direction coefficients $m=9, n=5, p=1$ and direction vector $\bar{s}=\{9 ; 5 ; 1\}$ or $\bar{s}=9 \bar{i}+5 \bar{j}+\bar{k}$.

The direction cosines of the line can be found using formulas (7):

$$
\begin{gathered}
\cos \alpha=\frac{m}{s}=\frac{9}{\sqrt{81+25+1}}=\frac{9}{\sqrt{107}} \\
\cos \beta=\frac{n}{s}=\frac{5}{\sqrt{107}} ; \cos \gamma=\frac{1}{\sqrt{107}}
\end{gathered}
$$