4.3. Form the equation of the plane passing through the points $P_{0}(2,-1,2), P_{1}(4,3,0), P_{2}(5,2,1)$.
Solution. Let's form the coordinates of the current vector of the plane $\overline{P_{0}} M=(x-2, y+1, z-2)$ and the vectors

$$
\overrightarrow{P_{0} P_{1}}=(2,4,-2), \overline{P_{0} P_{2}}=(3,3,-1) .
$$

For these vectors, the condition of coplanarity is valid:

$$
\left|\begin{array}{ccc}
x-2 & y+1 & z-2 \\
2 & 4 & -2 \\
3 & 3 & -1
\end{array}\right|=0
$$

We will bring the equation of the plane to the general form, for which we will calculate the determinant, using the rule of expansion by the elements of the first row:

$$
\begin{gathered}
\left|\begin{array}{ccc}
x-2 & y+1 & z-2 \\
2 & 4 & -2 \\
3 & 3 & -1
\end{array}\right|=(x-2)\left|\begin{array}{cc}
4 & -2 \\
3 & -1
\end{array}\right|-(y+1)\left|\begin{array}{cc}
2 & -2 \\
3 & -1
\end{array}\right|+(z-2)\left|\begin{array}{cc}
2 & 4 \\
3 & 3
\end{array}\right|= \\
=2 x-4 y-6 z+4=0, \quad x-2 y-3 z+2=0
\end{gathered}
$$