The line $CF$ is perpendicular to the plane $ABD$, so $CF \perp AD$. Similarly, $BE \perp AD$. Therefore, the lines $CF$ and $BE$ are parallel to the plane $\alpha$ or lie in it. The points $B$, $C$, $E$, and $F$ lie on the sphere $\omega$ circumscribed about the tetrahedron $ABCD$. Also, since $\angle BEC = 90^\circ = \angle BFC$, the points $B$, $C$, $E$, and $F$ lie on the sphere $\omega'$, constructed on the segment $BC$ as a diameter.

If the spheres $\omega$ and $\omega'$ do not coincide, all their common points lie in one plane, denote it by $\beta$. In the plane $\beta$ lie the lines $BE$ and $CF$, each of which is parallel to the plane $\alpha$ or lies in this plane. Also, the lines $BE$ and $CF$ are not parallel, since they are perpendicular to the intersecting planes $ACD$ and $ABD$. Thus, the plane $\beta$ is parallel to the plane $\alpha$ or coincides with it, and the distances from the points $E$ and $F$ to $\alpha$ are equal to the distance between the planes $\alpha$ and $\beta$ (see Fig. 12).

If the spheres $\omega$ and $\omega'$ coincide, then their common center $M$ is the midpoint of the segment $BC$ and lies in the plane $\alpha$. Therefore, the distances from the points $B$ and $C$ to $\alpha$ are equal. Since the line $BE$ is parallel to $\alpha$, the distances from $B$ and $E$ to $\alpha$ are equal. Similarly, the distances from $C$ and $F$ to $\alpha$ are also equal, and then the points $E$ and $F$ are equidistant from $\alpha$ (see Fig. 13).