Example 2. For the direction vector of the line

$$
\left\{\begin{array}{l}
2 x-3 y-3 z+4=0 \\
x+2 y+z-5=0
\end{array}\right.
$$

find the direction cosines.
Solution. According to point $6^{\circ}$, we find the direction vector $\vec{l}$ of the given line

$$
\vec{l}=\vec{N}_{1} \times \vec{N}_{2}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
2 & -3 & -3 \\
1 & 2 & 1
\end{array}\right|=3 \vec{i}-5 \vec{j}+7 \vec{k}
$$

Find $|\vec{l}|=\sqrt{9+25+49}=\sqrt{83}$. Now (Chapter III)

$$
\cos \alpha=\frac{3}{\sqrt{83}}, \quad \cos \beta=-\frac{5}{\sqrt{83}}, \quad \cos \gamma=\frac{7}{\sqrt{83}}
$$