I have a system of equation equation describing a class of 3D surfaces:
$\begin{cases} \sqrt{x^2+y^2+z^2}+ax+by+cz+d=0 \\ a^2+b^2+c^2-1=0 \end{cases}$
I'm trying to prove that there can be no set of eight points that satisfy the system of equations and are arranged like the vertices of a cube with edge length $1$ (the cube can be in any orientation); I tried to prove this for the lower-dimensional cases
$\begin{cases} \sqrt{x^2+y^2}+ax+by+c=0 \\ a^2+b^2-1=0 \end{cases} \mbox{ (four vertices of a square)}$
$\begin{cases} \sqrt{x^2}+ax+b=0 \\ a^2=1 \end{cases} \mbox{ (two points one unit apart)}$
I was able to prove the one-dimensional case quite easily (for b=0 there are infinite such sets, for every other value there are none), but this didn't give me much insight in the other ones. I hope this was clear, any help would be greatly appreciated!
It's easy to show that the only quadric surfaces that pass through the vertices of the cube with vertices $(\pm 1/2, \pm 1/2, \pm 1/2)$ are $$r (x^2-1/4) + s (y^2-1/4) + t (z^2-1/4) = 0 \tag{1}$$ Your equation (after isolating the square root on one side and squaring both sides) becomes a quadric. So the question is whether it can be obtained from something of the form (1) by translation and rotation.
I'm assuming you meant $cz$ rather than $cx$. Then after a suitable rotation we can take $a=1, b=0, c=0$, which makes your quadric $y^2 + z^2 - 2 d x - d^2 = 0$. This is a straight line if $d=0$, otherwise a paraboloid. Neither of those describe (1).