Let \(M_{1}\) be the midpoint of the segment \(B C\) and \(M_{2}\) the midpoint of the square \(A B F E\). According to the converse of a part of the ray theorem, \(M_{1} M_{2} \| C E\). The plane parallel to \(C E\) through \(A\) and \(F\) passes through \(M_{2}\) and thus contains the line through \(M_{2}\) parallel to \(C E\). Therefore, it also passes through \(M_{1}\).

The part of the body that it cuts off from the cube \(A B C D E F G H\) is consequently the tetrahedron \(A B F M_{1}\). If we consider the triangle \(A B F\) as the base and \(B M_{1}\) as the height of this tetrahedron, it follows that its volume is

\[
V_{T}=\frac{1}{3} \cdot \frac{1}{2} a^{2} \cdot \frac{1}{2} a=\frac{1}{12} a^{3}
\]

Analogously, for the other two cuts, we also obtain tetrahedra with the volume \(V_{T}\), where no two of these tetrahedra have common interior points.

Since \(3 V_{T}=\frac{1}{4} a^{3}\), the volume of the remaining body \(V_{R}=\frac{3}{4} a^{3}\).

Taken from \([5]\)
Task 1 - 151041
Given a cube \(A B C D E F G H\) with edge length \(a\). Through the points \(A\) and \(F\), \(A\) and \(H\), as well as \(F\) and \(H\), three planar cuts are made such that they are each parallel to the space diagonal \(E C\). These cuts separate three partial bodies from the cube.

Calculate the volume \(V_{R}\) of the remaining body!