Given four points not lying in the same plane. How many planes are there from which all four points are equidistant?
Let $S$ be a plane such that the given points are equidistant from it. None of the four points can lie on $S$, and they cannot all be on the same side of $S$, as they would then be coplanar. Therefore, there will be 1 and 3 points, or 2 and 2 points on the two sides of $S$.

If we select three out of the four points, there will be exactly one plane that separates these three from the fourth and is equidistant from all four points. This is the plane parallel to the plane defined by the three selected points and bisects the distance of the fourth point from this plane. Since we can choose three points out of four in four different ways, there will be 4 such planes.

Next, let's select two out of the four points. These two and the other two define skew lines, which can be uniquely placed in parallel planes. There will be exactly one plane, the mid-parallel plane of these parallel planes, that is equidistant from all four points. We can find such a plane in $\frac{1}{2} \cdot \binom{4}{2} = 3$ different ways.

Therefore, 7 planes satisfy the conditions of the problem.

Based on the work of László Fábián (Dobóvár, Illyés Gy. Gymnasium III. grade).