4.8. Form the equation of the line passing through the point $P_{0}(4,-3,2)$ and parallel to the given line:

$$
\left\{\begin{array}{l}
2 x-y-z+1=0 \\
3 x+2 y+z-8=0
\end{array}\right.
$$
Solution. Let's determine the coordinates of the direction vector of the line by finding the vector product of the vectors perpendicular to the planes $2 x-y-z+1=0$ and $3 x+2 y+z-8=0$. We have $\vec{n}_{1}=(2,-1,-1), \vec{n}_{2}=(3,2,1)$ :

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

$\vec{a}=(1,-5,7)$ - the direction vector of the line given in the condition, and therefore, it is also the direction vector of the desired line. The equation of the desired line expresses the condition of collinearity of vectors $\vec{a}$ and $\overline{P_{0} M}$ :

$$
(x-4) / 1=(y+3) /(-5)=(z-2) / 7
$$