Is it possible to write out natural number coordinates of four three-dimensional points $\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d} \in \mathbb{N}^3$, with the following determinant zero?
$$\left\vert \begin{array}{ccc} a_x - d_x & a_y - d_y & a_z - d_z \\ b_x - d_x & b_y - d_y & b_z - d_z \\ c_x - d_x & c_y - d_y & c_z - d_z \end{array} \right\vert$$
REFINED:
Dihedral angle with any coordinate plane should not been the multiple of $\frac{\pi}{2}$.
Any 3 points should not lies on line.
I think this satisfies what you are looking for: $(3,2,1),(1,1,1),(2,1,0),(0,0,0)$ Your matrix is then $$\left\vert \begin{array}{ccc} 3&2&1 \\ 1&1&1 \\ 2&1&0 \end{array} \right\vert=0$$ as the bottom two rows add to the top one.