Example 1.32. Form the equation of the plane passing through the line

\[
\begin{aligned}
3 x+2 y+5 z+6 & =0 \\
x+4 y+3 z+4 & =0
\end{aligned}
\]

parallel to the line

\[
\frac{x-1}{3}=\frac{y-5}{2}=\frac{z+1}{-3}
\]
Solution. Let's form the equation of the pencil of planes passing through the first of the given lines:

$$
\alpha(3 x+2 y+5 z+6)+\beta(x+4 y+3 z+4)=0
$$

which we divide by $\alpha \neq 0$, and let $\beta / \alpha=\lambda$ :

$$
\begin{aligned}
& 3 x+2 y+5 z+6+\lambda(x+4 y+3 z+4)=0, \text { or } \\
& (3+\lambda) x+(2+4 \lambda) y+(5+3 \lambda) z+(6+4 \lambda)=0
\end{aligned}
$$

In this pencil, we need to choose a plane that is parallel to the 2nd given line. From the condition of parallelism of the plane and the line, we have:

$$
3(3+\lambda)+2(2+4 \lambda)-3(5+3 \lambda)=0
$$

From this, $\lambda=1$.

Substituting $\lambda=1$ into the equation of the pencil of planes, we get:

$$
(3+1) \cdot x+(2+4 \cdot 1) y+(5+3 \cdot 1) z+(6+4 \cdot 1)=0 \text {. }
$$

Then the desired equation of the plane will be:

$$
\begin{gathered}
4 x+6 y+8 z+10=0 \text { or } \\
2 x+3 y+4 z+5=0 .
\end{gathered}
$$