\section*{Problem 2 - 121232}

In space, four points \(P_{1}, P_{2}, P_{3}\) and \(P_{4}\) are given, which do not lie in the same plane. Determine the number of all those planes that are equidistant from these four points.
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Preliminary remark: If the four points do not lie in a plane, then not three of them can lie on a common line. (1)

We distinguish how many points lie on each side of such a plane E.

1. Case: All four points lie on the same side of E. This case is not possible, because otherwise there would be a plane parallel to E that contains all four points.

2. Case: Three points lie on one side of E, the fourth on the other. The three points on one side of E lie in a plane E', which is parallel to E, as they all have the same distance to E. This plane is uniquely determined due to (1).

Now consider the perpendicular 1 from the fourth point to E', so 1 is also perpendicular to E. E must therefore bisect 1 exactly, so that the distance of all four points to E is the same. Thus, E is uniquely determined. E is parallel to E' and passes through the midpoint of 1.

Since there are four selection options for which point is the one that is alone on one side of E, this results in four different planes.

3. Case: Two points lie on each side. These two pairs determine two lines \(g\) and \(h\). These lines do not intersect and are not parallel, otherwise all four points would lie in a plane. The lines are therefore skew to each other.

A plane that has the same distance to two points lying on the same side also has the same distance to the entire line through these two points.

The plane E must be parallel to the lines \(g\) and \(h\) and bisect the distance of both lines. Thus, E is uniquely determined.

Since there are three selection options for which point is together with \(P_{1}\) on one side of E, this results in three different planes.

In total, there are therefore seven planes that have the same distance from all four points.