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 $$\overrightarrow{v}= (2, 1, 0)$$.
See development and solution
Development:
If we equate the following determinant to $$0$$, using the point and the director vectors, we obtain the general equation: $$$\left|\begin{matrix}x-1 & -1 & 2 \\ y & 3 & 1 \\ z-3 & 2 & 0 \end{matrix} \right|=-(z-3)+4y-6(z-3)-2(x-1)=-2x+4y-7z+23=0$$$
Solution:
Therefore the general equation is $$-2x + 4y - 7z + 23 = 0$$