4.9. Form the equation of the plane passing through the line $(x-1) / 2=(y-3) / 4=z /(-1)$ and the point $P_{1}(1,5,2)$.
Solution. Let's write the condition of coplanarity of three vectors: the current vector $\overrightarrow{P_{0} M}=(x-1, y-3, z)$, the vector $\overrightarrow{P_{0} P_{1}}=(0,2,2)$; belonging to the plane, and the direction vector of the given line $\vec{a}=(2,4,-1)$, parallel to the given plane:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-3 & z \\
0 & 2 & 2 \\
2 & 4 & -1
\end{array}\right|=(x-1)\left|\begin{array}{cc}
2 & 2 \\
4 & -1
\end{array}\right|-(y-3)\left|\begin{array}{cc}
0 & 2 \\
2 & -1
\end{array}\right|+z\left|\begin{array}{cc}
0 & 2 \\
2 & 4
\end{array}\right|= \\
& \quad=-10(x-1)+4(y-3)-4 z=0, \quad 5 x-2 y+2 z+1=0
\end{aligned}
$$

- the desired general equation of the plane.