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.
Your are right, such plane is not unique. For example the planes $2x-y+3z=18$ and $5x+4y-z=-13$ pass through the point $(2,-5,3)$ and they are parallel to the given line.
More generally, through the given point, there is a unique line parallel to the given line, but then any plane through this second line is parallel to the given line.
Best Answer
You could first find a normal vector to the sought after plane.
Your sought after plane is parallel to the plane containing both of the vectors $L_1$ and $L_2$. So, a vector normal to your plane is any vector, $\bf n$, perpendicular to both $L_1$ and $L_2$.
There are at least two ways to find $\bf n$:
$\ \ \ 1)$ Find a nonzero solution ${\bf n}=(a,b,c)$ to the system of equations ${\bf n}\cdot L_1={ 0}\,; \ {\bf n}\cdot L_2={ 0}$.
or
$\ \ \ 2)$ Take the cross product of $L_1$ and $L_2$.
Once you've found $\bf n$, then you can use the formula giving the equation of the plane with normal vector $ (a,b,c)$ that passes through the point $(x_0,y_0,z_0)$: $$ a(x-x_0)+b(y-y_0)+c(z-z_0)=0. $$