General equation of a plane
Determine the general equation of a plane that goes through point $A = (1, 0, 3)$ and has as director vectors $\overrightarrow{u}= (-1, 3, 2)$ and $\overrightarrow{v}= (2, 1, 0)$.
If we equate the following determinant to $0$, using the point and the director vectors, we obtain the general equation: $$\left|\begin{matrix}x-1 & -1 & 2 \\ y & 3 & 1 \\ z-3 & 2 & 0 \end{matrix} \right|=-(z-3)+4y-6(z-3)-2(x-1)=-2x+4y-7z+23=0$$
Therefore the general equation is $-2x + 4y - 7z + 23 = 0$