Converse of Parametric equation of plane Suppose a plane passes through a point A ( whose position vector is a ) and parallel to two vectors b and c . Then , to any general point on plane with position vector r , I can find the equation of plane as :
r = a + mb+ nc
for some scalars m and n , because to any general point r , the distance between a and r is a linear combination of b and c . Everything is fine and clear .
However, I am unable to prove it's converse i.e. if any plane written like
r = a + mb + nc
must pass through point with position vector a and must be parallel to vectors b and c . Please tell or atleast hint that how to prove it . Thanks in advance.
 A: The locus of points $\mathbf{p}$,
$$\mathbf{p} = \mathbf{a} + m \mathbf{b} + n \mathbf{c}, \quad m, n \in \mathbb{R} \tag{1}\label{1}$$
assuming $$\mathbf{b} \not\parallel \mathbf{c}, \quad \left\lVert \mathbf{b} \right\rVert \gt 0, \quad \text{and} \quad \left\lVert \mathbf{c} \right\rVert \gt 0$$
i.e. the two vectors $\mathbf{b}$ and $\mathbf{c}$ are nonzero and not parallel, by definition corresponds to plane
$$\mathbf{p} \cdot \mathbf{r} - d = 0 \tag{2}\label{2}$$
where
$$\left\lbrace \begin{aligned}
\mathbf{r} &= \mathbf{b} \times \mathbf{c} \\
d &= \mathbf{r} \cdot \mathbf{a} \\
\end{aligned} \right.$$
$\times$ denoting vector cross product and $\cdot$ vector dot product.
This is because equation $\eqref{1}$ is the parametric definition of a plane, and equation $\eqref{2}$ is the implicit definition of a plane, so by choosing the correct constants for the implicit definition, the two describe the exact same plane.
How to prove this?  I am not a mathematician, but perhaps a direct algebraic proof suffices.
Substituting $\mathbf{r}$ and $d$ from equation $\eqref{2}$ to equation $\eqref{1}$ yields
$$\begin{aligned}
\left(\mathbf{a} + m \mathbf{b} + n \mathbf{c} \right) \cdot \mathbf{r} -  \mathbf{a} \cdot \mathbf{r} &= 0 \\
\iff \quad \mathbf{a} \cdot \mathbf{r} + m \mathbf{b} \cdot \mathbf{r} + n \mathbf{c} \cdot \mathbf{r} - \mathbf{a} \cdot \mathbf{r} &= 0 \\
\iff \quad m \mathbf{b} \cdot \mathbf{r} + n \mathbf{c} \cdot \mathbf{r} &= 0 \\
\end{aligned} \tag{*}\label{*}$$
Due to the properties of vector cross product, $\mathbf{r} \perp \mathbf{b}$ and $\mathbf{r} \perp \mathbf{c}$; i.e. $\mathbf{r} \cdot \mathbf{b} = 0$ and $\mathbf{r} \cdot \mathbf{c} = 0$.  Therefore, for all real values of $m$ and $n$, equation $\eqref{*}$ is true, and the locus of points $\mathbf{p}$ describes the aforementioned plane.
Hopefully some mathematician member here can tell us whether this suffices as a proof, and how to express the idea more concisely ("more elegant proof", as they say).
