If we parametrize the plane $P:x+y+z=0$ (bijectively) using
then the orthogonal projection of the hexagon $H$
at the intersection of the cube $C$
onto $P$ can be found by simply describing $(u,v)$ along $H$.
(In fact $H$ is already in the plane $P$.)
In particular, if a symmetry about $(0,0)$ can be observed,
then we will have demonstrated that the centers of $C$ and $H$ coincide.
Using the given parametrization actually already guarantees that $(u,v)$
that $(x,y,z)$ lies in the plane $P$,
so we need only transform the equation for $C$:
This can be visualized in the $uv$-plane
as the hexagonal curve with the origin at its interior
connecting the finite line segments
of the six lines with equations
u+v&=&\pm s \\\\
u-v&=&\pm s \\\\
It's pretty easy to see that its center is $(0,0)$.
The three pairs of equations above are plotted
below in red, green and blue, respectively,
and the projections of the portions of the
$xzy$-coordinate axes interior to $C$
(between the centers of opposite faces)
is shown dotted in gray.