The coefficients of the variables give you the coordinates of a pair $v_1,v_2$ of vectors that are perpendicular to each plane. In you example the vectors $(1,2,-2)$ and $(0,1,0)$.
If the vectors were proportional it would mean that the planes are parallel and therefore the set of points equidistant from the two planes is the parallel plane in between. The quation of this plane would be $v_1\cdot (x,y,z)=(a+b)/2$ where $v_1\cdot(x,y,z)=a$ and $v_2\cdot(x,y,z)=b$ are the equations of the two given planes.
In your case the two vectors are not proportional. Therefore the planes intersect. In this case the set of points equidistant to the two planes are two planes that bisect the angle in between the two given planes.
To get these planes you just need the normal vectors to them. The displacement can then be found be imposing that they pass through some point of the intersection of the two given planes.
The normal vectors that we need are not the cross product of $v_1,v_2$. Such a cross product would point parallel to the line of intersection of the two planes.
What we need is the vectors that bisect the angles between $v_1$ and $v_2$. This bisector is easier to get if we first normalize $v_1$ and $v_2$, because then we just need to bisect a rhombus and the diagonals do this. We get $\frac{v_1}{\|v_1\|}+\frac{v_2}{\|v_2\|}$ and $\frac{v_1}{\|v_1\|}-\frac{v_2}{\|v_2\|}$ as the vectors we are looking for.
Find two points $P$ and $Q$ on the line of intersection of the planes $2x−3y−z+1=0$ and $3x+5y−4z+2=0.$
Then with $A=(3,−1,2)$ construct vectors $AP$ and $AQ$
The normal vector to your plane is the common perpendicular of $AP$ and $AQ$
Having a point and the normal vector you can easily find the equation of the plane.
Best Answer
Yes, you could convert equation (i) to parametric vector form, substitute it into (ii), make one parameter the subject, substitute it back into that parametric vector equation to finally obtain the intersection line's parametric vector equation. Notice that this doesn't actually involve finding $\vec a,$ which can be determined as the next step.
A more direct method (also not requiring finding $\vec a)$ is illustrated by this example: \begin{gather}\vec{r} \cdot \begin{pmatrix} 7\cr2 \cr -3\end{pmatrix}=4 \tag{$\pi_1$}\\\vec{r} \cdot \begin{pmatrix} 2\cr1 \cr 0\end{pmatrix}=5\tag{$\pi_2$} \end{gather}
$\pi_1:\;7x+2y-3z=4\\\pi_2:\;2x+y=5$
Expressing $x$ and $y$ in terms of $z$ $(\pi_1{-}2\pi_2,\,\ldots)$ gives the intersection line of $\pi_1$ and $\pi_2: \\\vec r=\begin{pmatrix} x\cr y \cr z\end{pmatrix}\\=\begin{pmatrix} z-2\cr 9-2z \cr z\end{pmatrix}\\=\begin{pmatrix} -2\cr 9 \cr 0\end{pmatrix}+z\begin{pmatrix} 1 \cr -2 \cr 1\end{pmatrix}\quad\left(z\in\mathbb R\right).$