[ $\underline{\text { equation of a plane ] }}$  [ Scalar product ]

Form the equation of the plane containing the line $\frac{x-1}{2}=-\frac{y}{3}=3-z$ and parallel to the line of intersection of the planes $4 x+5 z-3=0$ and $2 x+y+2 z=0$.
Vector $\overrightarrow{m_{1}}=(2 ;-3 ;-1)$ is the direction vector of the first line. Let $x=t$ and solve the system of equations for $y$ and $z$

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

We get

$$
z=-\frac{4}{5} t+\frac{3}{5}, y=-2 z-2 t=\frac{5}{5} t-\frac{6}{5}-2 t=-\frac{2}{5} t-\frac{6}{5}
$$

Thus, the direction vector of the line of intersection of the planes $4 x+5 z-3=0$ and $2 x+y+2 z=0$ can be taken as $\overrightarrow{m_{2}}=(5 ;-4 ;-2)$. Let $\vec{n}=(a ; b ; c)$ be a non-zero vector perpendicular to each of the two given lines. Then

$$
\left\{\begin{array}{l}
\vec{n} \cdot \overrightarrow{m_{1}}=0 \\
\vec{n} \cdot \overrightarrow{m_{2}}=0
\end{array}\right.
$$

or

$$
\left\{\begin{array}{l}
2 a-3 b-c=0 \\
5 a-4 b-2 c=0
\end{array}\right.
$$

Multiply the first equation by -2 and add it term by term to the second. We get $a+2 b=0$. Let $b=-1$. Then $a=2, c=2 a-3 b=7$. Thus, a vector perpendicular to the desired plane can be taken as $\vec{n}=(2 ;-1 ; 7)$. Through an arbitrary point on the line $\frac{x-1}{2}=-\frac{y}{3}=3-z$, for example, through the point $P(1 ; 0 ; 3)$, we draw a plane perpendicular to the vector $\vec{n}=(2 ;-1 ; 7)$ :

$$
2(x-1)-y+7(z-3)=0, \text { or } 2 x-y-7 z-23=0 .
$$

## Answer

$2 x-y+7 z-23=0$.