Solution:

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.