413. Find the angle between the line

\[
\left\{\begin{array}{l}
x-2 z+3=0 \\
y+3 z-1=0
\end{array}\right.
\]

and the plane

\[
2 x-y+z+3=0
\]
Solution. We will write the equations of the line (A) in canonical form. From the first equation of the system (A) $z=\frac{x+3}{2}$, from the second equation $z=\frac{1-y}{3}$, hence

$$
\frac{x+3}{2}=\frac{1-y}{3}=\frac{z}{1}
$$

or

$$
\frac{x+3}{2}=\frac{y-1}{-3}=\frac{z}{1}
$$

Therefore, the direction coefficients of the line are $m=2$, $n=-3$, $p=1$, and by the condition, the coordinates of the normal vector of the plane (B) are $A=2$, $B=-1$, $C=1$. To determine the angle, we will use formula (12):

$$
\sin \varphi=\frac{2 \cdot 2+(-1) \cdot(-3)+1 \cdot 1}{\sqrt{4+1+1} \sqrt{4+9+1}}=\frac{4 \sqrt{21}}{21} \approx 0.8730
$$

This value of $\sin \varphi$ corresponds to the angle $\varphi \approx 60^{\circ} 49^{\prime}$.