Find the angle between the line of intersection of the planes $2 x-y-3 z+5=0$ and $x+y-2=0$ and the plane passing through the points $M(-2 ; 0 ; 3), N(0 ; 2 ; 2)$ and $K(3 ;-3 ; 1)$.

#
First, let's find the direction vector of the line of intersection of the given planes. For this, let $x=t$ and solve the system of equations for $y$ and $z$:

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

We get

$$
x=t, y=2-t, z=1+t
$$

Thus, the vector $\overrightarrow{\boldsymbol{m}}=(1 ;-1 ; 1)$ is parallel to the line of intersection of the given planes. Next, we need to find a vector perpendicular to the plane passing through the points $M(-2 ; 0 ; 3), N(0 ; 2 ; 2)$, and $K(3 ;-3 ; 1)$. We will compute the coordinates of the vectors $\overrightarrow{M N}$ and $\overrightarrow{M K}$:

$$
\begin{aligned}
& \overrightarrow{M N}=(0-(-2) ; 2-0 ; 2-3)=(2 ; 2 ;-1) \\
& \overrightarrow{M K}=(3-(-2) ;-3-0 ; 1-3)=(5 ;-3 ;-2)
\end{aligned}
$$

Let $\vec{n}=(a ; b ; c)$ be a non-zero vector perpendicular to the desired plane. Then $\vec{n} \cdot \overrightarrow{M N}=0$ and $\vec{n} \cdot \overrightarrow{M K}=0$ or

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

Multiply both sides of the first equation by -2 and add the result term by term to the second equation. We get the equation $a-7 b=0$. Let $b=1$. Then $a=7, c=2 a+2 b=16$. Let $\phi$ be the angle between the vectors $\vec{m}=(1 ;-1 ; 1)$ and $\vec{n}=(7 ; 1 ; 16)$. Then

$$
\begin{aligned}
& \cos \phi=\frac{\vec{m} \cdot \vec{m}}{|\vec{m}| \cdot|\overrightarrow{\mid \vec{k}}|}=\frac{1 \cdot 7-1 \cdot 1+1 \cdot 16}{\sqrt{1^{2}+1^{2}+1^{2} \cdot \sqrt{7^{2}+1^{2}+16^{2}}}}= \\
& =\frac{22}{\sqrt{3 \cdot 306}}=\frac{22}{3 \sqrt{102}}
\end{aligned}
$$

If $\alpha$ is the angle between the specified line and the plane $M N K$, then

$$
\sin \alpha=|\cos \phi|=\frac{22}{3 \sqrt{102}}
$$

## Answer

$\arcsin \frac{22}{3 \sqrt{102}}$