4.11. Form the equation of the plane equidistant from the points $B(4,1,0)$ and $C(2,0,3)$.
Solved. We will use the distance formula between points to find the coordinates of point $A(0,0, z)$, equidistant from the two given points, and set the magnitudes $|A B|=|A C|$.

We have

$$
\begin{aligned}
& |A B|=\sqrt{(4-0)^{2}+(1-0)^{2}+(0-z)^{2}} \\
& |A C|=\sqrt{(2-0)^{2}+(0-0)^{2}+(3-z)^{2}} \\
& 17+z^{2}=4+(3-z)^{2}, \quad z=-2 / 3
\end{aligned}
$$

The equation of the desired plane will be obtained from the condition of orthogonality of the current vector of the plane $\overline{A M}=(x, y, z+2 / 3)$ and the vector $\overrightarrow{B C}=(-2,-1,3)$:

$-2 \cdot x-1 \cdot y+3 \cdot(z+2 / 3)=0,-2 x-y+3 z+2=0$.