Prove that there is no inclined plane such that any tetrahedron placed arbitrarily with a certain face on the plane will not fall over. It means the following:Given a plane π and a line l not perpendicular to it. Prove that there is a tetrahedron T such that for each of its faces S there is in the plane π a triangle ABC congruent to S and there is a point D such that the tetrahedron ABCD congruent to T and the line parallel to l passing through the center of gravity of the tetrahedron ABCD does not intersect the triangle ABC.Note. The center of gravity of a tetrahedron is the intersection point of the segments connecting the centers of gravity of the faces of this tetrahedron with the opposite vertices (it is known that such a point always exists). 3D geometrygeometrytetrahedron