Position the plane such that it intersects the cube in a regular hexagon going through the centers of six of the twelve edges of the large cube, equidistant from two of the cube's vertices that are opposite ends of a cube diagonal. You can model it with a rubber band around this "equator" going through the midpoint of each of the six edges it goes through.

A similar hexagon is intercepted by the central cubelet, as the intersection of the same plane with that cubelet. .

The large model also models the small center cubelet. The rubber band goes through each of the faces, which correspond to each face of the center cube, sharing that face with an adjacent cube, which is the center face cubelet of the larger, rubik's, cube. So the 6 center faces are all intersected.

From the rubber band you can also see that 6 edge and 6 corner pieces are intersected.

That makes 1 center cubelet, 6 face-center cubelets, 6 edge cubelets (out of twelve) and 6 corner cubelets (out of eight), for a total of 19.

Is this the best that can be done?

If the hexagon about the central cubelet is drawn on paper, the extended lines of that hexagon are the intersections of the plane with the interfaces of the cubelets

The triangles formed on the star of David are the intersections of the plane with the six cubelets that share a face with the central cubelet--these are the cube face centers--there are 6 of them.

From the traces on the surface, we can also see that 6 edge pieces and 6 corner pieces are intercepted by the plane.

The total is 1+6+6+6 = 19. These correspond to the 19 regions into which the six pairs of parallel lines divide the plane. That shows that no other plane will cut more cubelets.

No other way of drawing three pairs of parallel lines will cut more regions of the plane.