If for example my plane has a vector equation of $$\mathbf{r}=(3,1,3)+e(3,1,7)=m(3,-1,4),$$ where $e$ and $m$ are real parameters, how do I transform this parametric form to a scalar product form of $$\mathbf{r}\cdot\mathbf{n}=\mathbf{a}\cdot\mathbf{n}\;?$$
I know $$\mathbf{n}= (3,1,7) \times (3,-1,4),$$ but is my $\mathbf{a}$ (known point) $(3,1,3)$, or either $(3,1,7)$ or $(3,-1,4)$ ?
Using vector algebra, a 3D plane is often defined using is normal vector $\vec{n}$ and signed distance from origin (in units of normal vector length) $d$, such that point $\vec{p}$ is on the plane if and only if $$\vec{p} \cdot \vec{n} = d$$
If you have three non-collinear points (not on the same line) on the plane, $\vec{v}_1$, $\vec{v}_2$, and $\vec{v}_3$, you can calculate $\vec{n}$ and $d$ using $$\begin{aligned} \vec{n} &= \left( \vec{v}_2 - \vec{v}_1 \right) \times \left( \vec{v}_3 - \vec{v}_1 \right ) \\ d &= \vec{n}\cdot\vec{v}_1 = \vec{n}\cdot\vec{v}_2 = \vec{n}\cdot\vec{v}_3 \\ \end{aligned}$$ Note that you can scale $\vec{n}$ by any nonzero scalar value $\lambda$ (i.e., multiply each component of $\vec{n}$ by $\lambda$) as long as you also multiply $d$ by that same value, without affecting the plane.
While the above defines the plane, it does not identify, or parametrise, the points on the plane.
We can identify or parametrise the points on a plane if we have three vectors: $\vec{p}_0$ specifying the origin of the plane (the point we'll call $(0,0)$ in our plane coordinate system), $\vec{p}_u$ our unit $u$ unit vector, and $\vec{p}_v$ our unit $v$ axis vector. Then, point $(u, v)$ on the plane corresponds to point $\vec{p}$ in the 3D coordinates: $$\vec{p} = \vec{p}_0 + u \vec{p}_u + v \vec{p}_v \tag{1}\label{AC1}$$
If we use $$\vec{p} = \left [ \begin{matrix} x \\ y \\ z \end{matrix} \right ], \quad \vec{p}_0 = \left [ \begin{matrix} x_0 \\ y_0 \\ z_0 \end{matrix} \right ], \quad \vec{p}_u = \left [ \begin{matrix} x_u \\ y_u \\ z_u \end{matrix} \right ], \quad \vec{p}_v = \left [ \begin{matrix} x_v \\ y_v \\ z_v \end{matrix} \right ]$$ then we can write $\eqref{AC1}$ as a system of three scalar equations: $$\left\lbrace ~ \begin{aligned} x &= x_0 + u x_u + v x_v \\ y &= y_0 + u y_u + v y_v \\ z &= z_0 + u z_u + v z_v \\ \end{aligned} \right . \tag{2}\label{AC2}$$ If we want the inverse, i.e. to know the $(u, v)$ coordinates of a point $\vec{p}$ on a plane defined using the three vectors $\vec{p}_0$, $\vec{p}_u$, and $\vec{p}_v$, we need to realize we have three equations but only two unknowns.
Simply put, we need to drop the least important Cartesian coordinate axis. To do this, we calculate $\vec{n} = \vec{p}_u \times \vec{p}_v$, and drop the coordinate axis corresponding to the largest component in magnitude of $\vec{n}$.
Let's say that component is $z$, i.e. the plane is not too far tilted compared to the $xy$ plane. Then, we solve $$\left\lbrace ~ \begin{aligned} x &= x_0 + u x_u + v x_v \\ y &= y_0 + u y_u + v y_v \\ \end{aligned} \right . \tag{3.a}\label{AC3a}$$ for $u$ and $v$. $$\left\lbrace ~ \begin{aligned} u &= \frac{(x - x_0) y_v - (y - y_0) x_v}{x_u y_v - x_v y_u} \\ v &= \frac{(y - y_0) x_u - (x - x_0) y_u}{x_u y_v - x_v y_u} \\ \end{aligned} \right . \tag{3.b}\label{AC3b}$$ Similarly for the two other possibilities, dropping the $y$ or $x$ coordinates.
If we look at OP's definition, $$\vec{r} = \vec{a} + e \vec{b} = m \vec{c}$$ where $$\vec{a} = \left [ \begin{matrix} 3 \\ 1 \\ 3 \end{matrix} \right ], \quad \vec{b} = \left [ \begin{matrix} 3 \\ 1 \\ 7 \end{matrix} \right ], \quad \vec{c} = \left [ \begin{matrix} 3 \\ -1 \\ 4 \end{matrix} \right ], \quad e \in \mathbb{R}, \quad m \in \mathbb{R}$$ In Cartesian coordinates, this is $$\left\lbrace ~ \begin{aligned} x &= 3 + 3 e = 3 m \\ y &= 1 + 1 e = -1 m \\ z &= 3 + 7 e = 4 m \\ \end{aligned} \right . $$ but this has no solutions for $e \in \mathbb{R}$, $m \in \mathbb{R}$, so the equation OP shows defines nothing: it is never true.
However, if the second $=$ is actually supposed to be a $+$ (or a $-$), then we'd have the situation I explained before, with $\vec{p}_0 = (3, 1, 3)$, $\vec{p}_e = (3, 1, 7)$, and $\vec{p}_m = (3, -1, 4)$, except using $(e, m)$ instead of $(u, v)$. Then, the equation would specify a vector-valued function $\vec{r}(e, m)$ that defines the location of point $(e, m)$ on the plane.
The normal of that plane would be $\vec{n} = \vec{p}_e \times \vec{p}_m = (11, 9, -6)$, and the signed distance to origin (in normal vector units) $d = \vec{n} \cdot \vec{p}_0 = 24$, giving the equation of this plane (as $\vec{n}\cdot\vec{p} = d$).
If we normalize $\vec{n}$ to unit length (by multiplying each component by $1/\sqrt{\vec{n}\cdot\vec{n}}$), the unit normal is $(11/\sqrt{238}, 9/\sqrt{238}, -6/\sqrt{238}) \approx (0.713024, 0.583383, -0.388922)$, and the actual distance from origin is $d = 24/\sqrt{238} \approx 1.555689$.
As to the inverse solution -- finding the $e$ and $m$ that correspond to some point $(x, y, z)$ on this plane -- it is best to drop $x$ coordinates (because the $x$ component of $\vec{n}$ is the largest in magnitude of the three), and only use $y$ and $z$ for solving $e$ and $m$. In this case, the inverse solution is $$\left\lbrace ~ \begin{aligned} \displaystyle e &= \frac{4 y + z - 7}{11} \\ \displaystyle m &= \frac{z - 7 y + 4}{11} \\ \end{aligned} \right.$$
So, while I am not sure if this is anything like what the OP asked for, I hope this will answer any questions others might have, when looking for help on a question that matches the Subject.