# Four coplanar points in $\mathbb{N}^3$ space

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.

• Sure: $(1,0,0),(2,0,0),(3,0,0),(4,0,0)$. – vadim123 Jun 28 '14 at 18:40
• @vadim123 Mmmm). Small refinement: dihedral angle with any coordinate plane is not multiple of $\frac{\pi}{2}$. – Orient Jun 28 '14 at 19:06
• Still sure: $(1,1,1),(2,2,2),(3,3,3),(4,4,4)$ – Ross Millikan Jun 28 '14 at 19:14
• @RossMillikan what about the last statement? (Sorry for plenty of editions, but I could not formulate the problem immediately). – Orient Jun 28 '14 at 20:25
• How about $(0,0,0),(1,0,1),(0,1,1),(1,1,2)$? – Ross Millikan Jun 28 '14 at 22:06

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.
• Whole numbers in $\mathbb{N}$ are strictly positive things. – Orient Jun 29 '14 at 5:32
• You can add one to each coordinate of each point, producing the same matrix. It should be clear how to make such a matrix by now. Make your favorite $3 \times 3$ singular matrix. Let $d=(1,1,1)$ Solve for the other three points. Done. – Ross Millikan Jun 29 '14 at 14:23