Here is one way:
Line $p_1$ is defined by $(1,2,3)x+(0,-1,2)$, while $p_2$ is defined by $(1,-1,4)x+(0,1,-1)$.
So the direction vector for $p1$ is $(1,2,3)$ and the direction vector for $p2$ is $(1,-1,4)$. A quick way to find a vector perpendicular to both of them is to calculate their cross product, which is $(11,-1,-3)$.
Therefore any plane that is perpendicular to the vector $(11,-1,-3)$ will be parallel to or coincident with your two given lines. The equation of such a plane is
$$11x-y-3z=c$$
for some constant $c$.
If the plane is equidistant from the two lines, then the value of $c$ will be equidistant from the values of that expression when a point on each line is substituted. The value of $11x-y-3z$ for point $(0,-1,2)$ on line $p_1$ is $-5$, while the value for point $(0,1,-1)$ on line $p_2$ is $2$.
Therefore we want $c$ to be the average of $-5$ and $2$, which is $-\frac 32$. Your desired plane is
$$11x-y-3z=-\frac 32$$
Yes, the vector equation and the coordinate equation for a plane both represent the same thing. Often times, we use vectors to find the equation of a plane simply because doing it that way is the easiest. I'll explain the connection between the two.
Say that we fix a point $P_0 = (x_0,y_0,z_0)$ on some plane, and we find a vector $\textbf{n} = \langle a,b,c\rangle$ that is normal to the plane as you can see here at Wolfram Mathworld. To find the equation of the plane, what we would like is to be able to describe any other point $P = (x,y,z)$ on the plane. We can do this in a very simple way using vector manipulations.
Say that $\textbf{r}$ is the position vector of point $P$ (the vector going from the origin to point $P$), and $\textbf{r}_0$ is the position vector of point $P_0$. Then, the vector $\textbf{r} - \textbf{r}_0$ goes from point $P_0$ to point $P$. Now, since planes are "flat" we know that if point $P$ is to be on the plane, the vector $\textbf{r} - \textbf{r}_0$ must be perpendicular to the vector $\textbf{n}$. This means that
$$
\textbf{n} \cdot (\textbf{r} - \textbf{r}_0
) = 0 \tag{1}
$$
This is the equation of the plane in vector form. Often, this is not the end result you are looking for. Because you want everything in terms of coordinates. But it is not too hard to get to the form we want. Taking the vector equation and plugging in the coordinates we have
$$
\begin{align}
0 &=\langle a,b,c \rangle \cdot \left[ \langle x,y,z \rangle - \langle x_0,y_0,z_0 \rangle \right] \\ &= \langle a,b,c \rangle \cdot \langle x - x_0,y-y_0,z-z_0 \rangle \\ & = a(x-x_0) + b(y - y_0) + c(z - z_0)
\end{align}
$$
And so at last, here is the coordinate equation:
$$
a(x-x_0) + b(y - y_0) + c(z - z_0) = 0 \tag{2}
$$
Sometimes this is multiplied out, and all of the non-$x,y,z$ terms are collected on the right hand side and given a name like $D$. That makes the equation look like this
$$
ax + by + cz = D \tag{3}
$$
where $D = ax_0 + by_0 + cz_0$. So, you now have a few equivalent forms to work with, and what you choose to use depends entirely on the information you are given. For instance, if you are asked to find the equation of the plane, passing through the point $P_0 = (1, 2, 3)$ normal to the vector $\textbf{n} = \langle 5, 4, 7 \rangle$ you can immediately use $(2)$ to write $5(x - 1) + 4(y - 2) + 7(z - 3) = 0$. On the other hand, if you are asked, for instance, to find a vector normal to the plane $3x + y + 2z = 24$ you can deduce from $(3)$ immediately that $\textbf{n} = \langle 3,1,2 \rangle$.
Best Answer
An operator that transports vectors from one plane to another is sometimes called a “discrete connection” and has been studied in the context of simplicial complexes etc.
One canonical way to transport a vector from one plane to another: rotate the first plane about the edge common to both planes until the first plane is parallel to the second. Then slide the vector onto the second plane.