It is not possible. Consider $P$, the projection of \left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$onto the plane. Since for any point$(x, y, z)$in the cube,$(1-x, 1-y, 1-z)$is also in the cube, and the midpoint of their projections will be the projection of their midpoint, which is$P$, the projection of the cube onto this plane will be a centrally symmetric region around$P$, and thus cannot be a pentagon.