\section*{Problem 3 - 261013}

Imagine three planes (not necessarily distinct from each other) passing through the center of a sphere.

Into how many regions can the surface of the sphere be divided by such planes? Consider different cases to determine all possible numbers of regions!
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There are exactly the following cases for three planes through the center \(M\) of a sphere:

1. Case: All three planes are identical.

In this case, exactly two partial surfaces are created (the two hemispherical surfaces, whose common boundary is the great circle in which the plane intersects the spherical surface).

2. Case: Exactly two of the three planes are identical.

In this case, the two great circles created by the intersection of the non-identical plane with the spherical surface intersect at exactly two points \(P\) and \(Q\).

Each of the two hemispherical surfaces into which the spherical surface is divided by one great circle is further divided into two partial surfaces by the second circle. Therefore, a total of four partial surfaces are created.

3. Case: No two of the three planes are identical.

In this case, initially, two of the planes create two great circles, which intersect at two points \(P, Q\) and divide the spherical surface into four partial surfaces, as in case 2. The line \(g\) through \(P\) and \(Q\) contains the center \(M\) of the sphere, as it is the intersection line of the two planes and these pass through \(M\).

Now there are exactly the following two sub-cases:

3.1. The line \(g\) is also contained in the third plane. Then the third intersection circle passes through \(P\) and \(Q\). Exactly two of the four partial surfaces already created are further divided. Thus, exactly six partial surfaces are created.

3.2. The line \(g\) is not contained in the third plane. Then the third intersection circle does not pass through \(P\) or \(Q\). It intersects the other two intersection circles at two points each, different from \(P\) and \(Q\). This results in a total of four points, which divide this third circle into four segments. Each of these four segments acts as a new boundary for two new partial surfaces that have emerged from an old one. Therefore, a total of \(2 \cdot 4 = 8\) partial surfaces are created.

The sought-after possibilities are therefore: Two, four, six, or eight partial surfaces can be created.