The points on the line of intersection $\cal L$ of $\pi$ and the $xz$-plane satisfy $x + 2(0) - 2z + 3 = 0$, so $x - 2z = -3$. In particular, points $(0,0,\frac{3}{2})$, and $(1,0,2)$ are on the line, so $\cal L$ has direction vector $<1,0,\frac{1}{2}>$, or $<2,0,1>$. So the plane perpendicular to $\pi$ through $\cal L$ should have normal vector that is perpendicular to both $<1,2,-2>$ and $<2,0,1>$, and the plane passes through $(1,0,2)$, so this should be enough to finish up the problem.
Another condition "for a line to intersect another line, in 3D" is that there is a point that is on both lines. That sounds simplistic but that really is the condition.
Here is one way to solve your problem. The line that you want is in a plane that is parallel to the plane $3x-y+2z-15=0$. Any such plane has the equation
$$3x-y+2z+D=0$$
for some constant $D$. That plane must also go through the point $M(1,0,7)$, so we can substitute to find the value of $D$:
$$3\cdot 1-0+2\cdot 7+D=0$$
So $D=-17$ and the plane is $3x-y+3z-17=0$.
Combining that with your equations $\frac{x-1}{4}=\frac{y-3}{2}=\frac{z}{1}$ you can find the single point that is the intersection of that plane and your given line. The line you want goes through those two points.
You should be able to finish from here.
There are multiple ways to find the intersection of the plane and of the given line. You showed one way, by giving a parameterization of the line and solving for the parameter. You did make a computation error: the equation you get is
$$3(4t+1)-(2t+3)+2(t)-17=0$$
and the solution is
$$t=\frac{17}{12}$$
which makes $x=\frac{20}3,\ y=\frac{35}6,\ z=\frac{17}{12}$. Substituting this solution into all the equations checks, so this is the right answer. Thus the intersection point is
$$Q\left(\frac{20}3,\ \frac{35}6,\ \frac{17}{12}\right)$$
I got the intersection point by different means, by setting up and solving these simultaneous linear equations:
$$\begin{align}
\color{white}{1}x \color{white}{+0y}-4z&=1 \\
y-2z&=3 \\
3x-y+2z&=17
\end{align}$$
I got the same intersection point.
Best Answer
The plane you are looking for contains the normal vector to $\pi_1$, and also contains the direction vector of $l$. As you point out, those are $\langle 1, -3, -1\rangle$ and $\langle 2, -3, 1\rangle$. Can you use that information to figure out a normal vector to the plane you are looking for, then put that together with the given information about $P$ to find your answer?