Show that the equation for a plane through the points $(x_1,y_1,z_1),(x_2,y_2,z_2),(x_3,y_3,z_3)$ can be written as
$$\begin{vmatrix}x&y&z&1\\x_1&y_1&z_1&1\\x_2&y_2&z_2&1\\x_3&y_3&z_3&1\end{vmatrix}=0$$
Is there some clever way of doing this that is not just brute force calculations of the determinant and then calculate the norm vector for the plane, which will be the coefficients for x,y,z-values in the plane equation:
$$\begin{pmatrix}x_3-x_1\\y_3-y_1\\z_3-z_1\end{pmatrix}\times\begin{pmatrix}x_2-x_1\\y_2-y_1\\z_2-z_1\end{pmatrix}$$
and then calculate the equations constant by scalar multiply the norm vector with one of the points.
Best Answer
$$\begin{vmatrix}x&y&z&1\\x_1&y_1&z_1&1\\x_2&y_2&z_2&1\\x_3&y_3&z_3&1\end{vmatrix}=0$$ implies that the four columns $C_1,C_2,C_3,C_4$ are linearly dependent. And therefore that it exists $a,b,c,d$ non all equal to zero such that $aC_1 + bC_2+cC_3+dC_4=0$, i.e. that
$$ax+by+cz+d=0$$ is an equation of a plane containing the three initial points $P_1,P_2,P_3$.