Equation of a plane determined by a point and $2$ vectors

Question

I was taught that if $\pi$ is a plane, then it can be determined by a point $A_1(x_1,y_1,z_1)\in \pi$ and $\vec{u_k}=(l_k,m_k,n_k)$, $k=\overline{1,2}$ (non-null, non-collinear), using this formula $$\pi: \begin{vmatrix} x-x_1 & l_1 & l_2 \\ y-y_1 & m_1 & m_2 \\ z-z_1 & n_1 & n_2 \end{vmatrix}=0$$ I thought that the vectors $\vec{u_k}$ need to be in our plane $\pi$, but in an exercise they were just parallel to $\pi$ and the instructor used the same formula. Why does this work?

Answer 1

This works because, $(x,y,z)\in\pi$ if and only if $(x-x_1,y-y_2,z-z_3)$ is a linear combination of $\vec{u_1}$ and $\vec{u_2}$, that is, if and only if that determinant that you mentioned is $0$.