RU-ALLRUSSIAN 2021 Problem 116

In tetrahedron $ABCS$ no two edges have equal length. Point $A'$ in plane $BCS$ is symmetric to $S$ with respect to the perpendicular bisector of $BC$ . Points $B'$ and $C'$ are defined analagously. Prove that planes $ABC, AB'C', A'BC'$ abd $A'B'C$ share a common point.