374. Form the equation of the plane passing through the points $M(1 ; 2 ; 0), N(1 ; -1 ; 2), P(0 ; 1 ; -1)$ and find the angles of its normal with the coordinate axes.
Solution. It is known that the position of a plane is determined by three points (not lying on the same line). Let's write the equation of any plane passing through the point $M(1; 2; 0):$

$$
A(x-1)+B(y-2)+C z=0
$$

To obtain the desired equation of the plane, we need to require that the coordinates of points $N$ and $P$ satisfy equation $(A)$:

$$
\left\{\begin{array} { l } 
{ - 3 B + 2 C = 0 , } \\
{ - A - B - C = 0 }
\end{array} \quad \text { or } \left\{\begin{array}{l}
3 B-2 C=0 \\
A+B+C=0
\end{array}\right.\right.
$$

From here, $B=\frac{2}{3} C, A=-C-B=-\frac{5}{3} C$. Substituting these values into equation (A), we get the desired equation of the plane:

$$
-\frac{5}{3} C(x-1)+\frac{2}{3} C(y-2)+C z=0
$$

or

$$
\frac{5}{3}(x-1)-\frac{2}{3}(y-2)-z=0
$$

or

$$
5 x-2 y-3 z-1=0
$$

To determine the angles formed by the normal vector $\bar{n}\{5 ;-2 ;-3\}$ of the desired plane with the coordinate axes, we use formulas (14):

$$
\begin{aligned}
& \cos \alpha=\frac{5}{\sqrt{25+4+9}}=\frac{5 \sqrt{38}}{38} \approx 0.8111 \\
& \cos \beta=-\frac{2}{\sqrt{38}}=-\frac{\sqrt{38}}{19} \approx-0.3244 \\
& \cos \gamma=-\frac{3}{\sqrt{38}}=-\frac{3 \sqrt{38}}{38} \approx-0.4866
\end{aligned}
$$

From here, $\alpha \approx 35^{\circ} 48^{\prime}, \beta \approx 108^{\circ} 56^{\prime}, \gamma \approx 119^{\circ} 7^{\prime}$.