Not coplanar points in $\mathbb{R}^3$ if and only if its determinant is nonzero How to prove that points $P_1(x_1,y_1,z_1)$, $P_2(x_2,y_2,z_2)$,  $P_3(x_3,y_3,z_3)$, $P_4(x_4,y_4,z_4)$ are not coplanar if and only if the determinant  $$\left|\begin{array}{cccc}
                    x_1 & y_1 & z_1 & 1 \\
                    x_2 & y_2 & z_2 & 1 \\
                    x_3 & y_3 & z_3 & 1 \\
                    x_4 & y_4 & z_4 &1 \\
                  \end{array}
                \right| \neq 0$$
 A: If the points are coplanar, then you can find the equation of a plane $P \equiv ax+by+cz+d=0$ such that your points all verify the equation with $(a,b,c,d) \neq (0,0,0,0)$. Hence the system $$\begin{cases}
                    x_1 a + y_1 b + z_1 c + 1 d = 0 \\
                    x_2 a + y_2 b + z_2 c + 1 d = 0 \\
                    x_3 a + y_3 b + z_3 c + 1 d = 0 \\
                    x_4 a + y_4 b + z_4 c + 1 d = 0
                  \end{cases}$$
has a non trivial solution and the determinant vanishes (if not $(0,0,0,0)$ would be the only solution).
Conversely, if the determinant vanishes, the system has at least a non zero solution $(a,b,c,d)$ and this forms the equation of a plane satisfied by the points which are coplanar.
A: Claim:

The absolute value of this determinant
  $$\left|\begin{array}{cccc}
                     x_1 & y_1 & z_1 & 1 \\
                     x_2 & y_2 & z_2 & 1 \\
                     x_3 & y_3 & z_3 & 1 \\
                     x_4 & y_4 & z_4 &1 \\
                   \end{array}
                 \right| $$
  is the volume of a (skew) parallelepiped with vertices
  $(x_1,y_1,z_1),(x_2,y_2,z_2),(x_3,y_3,z_3),(x_4,y_4,z_4)$.

Proof:
We know that the volume is the absolute value of the determinat that has as columns the vectors that represents the sides of the parallelepiped:
$$ d=
\left|\begin{array}{cccc}
                     x_1-x_4 & y_1-y_4 & z_1-z_4 \\
                     x_2-x_4 & y_2-y_4 & z_2-z_4 \\
                     x_3-x_4 & y_3-y_4 & z_3-z_4\\                     
                   \end{array}
                 \right|
$$
Now, by co-factor decomposition of a determinant we have:
$$
d=
\left|\begin{array}{cccc}
                     x_1-x_4 & y_1-y_4 & z_1-z_4&1 \\
                     x_2-x_4 & y_2-y_4 & z_2-z_4&1 \\
                     x_3-x_4 & y_3-y_4 & z_3-z_4&1\\                     
                     0&0&0&1
\end{array}
                 \right|
$$
and, multiply the last columns bi $x_4$ and adding to the frst column we find
$$
d=
\left|\begin{array}{cccc}
                     x_1 & y_1-y_4 & z_1-z_4&1 \\
                     x_2 & y_2-y_4 & z_2-z_4&1 \\
                     x_3 & y_3-y_4 & z_3-z_4&1\\                     
                     x_4&0&0&1
\end{array}
                 \right|
$$
Doing the same,i.e. multiplying by $y_4$ and $z_4$ and adding to the second and third columns we arrive at:
$$
d=
\left|\begin{array}{cccc}
                     x_1 & y_1 & z_1&1 \\
                     x_2 & y_2 & z_2&1 \\
                     x_3 & y_3 & z_3&1\\                     
                     x_4&y_4&z_4&1
\end{array}
                 \right|
$$
That prove the claim.
Now the answer is obvious since the volume is null iff the four points are coplanar.
