Basically there won't really be one possible answer for this question, you can have to two planes that can be rotated by $90$ , the other not shown in the picture. To approach at this I think you will have to take these condition. Since I have had such a question but don't remember exact algorithm now.
1) The rotated plane contain the line. And thus any point on the line as well.
2) The plane is normal to the plane from which is it rotated. (This is the condition which will give two roots , hence two planes).
3)You can also try family of planes. You simply require a plane passing through the intersection of that plane.
This exact approach as far as i remember was quite messy.
$\color{red}{EDIT:}$ I just realized the two planes either by rotating $90$ clockwise or anticlockwise are actually equivalent, there will be only single plane as an answer. Here's why.
Pictorially, the two "distinct" should be $180$ from each other, but that's just equivalent to $0$ in case of planes. It's only a matter of orientation, and since these are nor vectors having defined normal vector. Both of them are equivalent.
2) Here's how to approach it algebraically.
Any plane through intersection of two planes is $(x+y+z-1)+\lambda(x-2y+3z)=0$
It is normal to the plane $(x+y+z-1)=0$ , which gives $\lambda=\frac{-3}{2}$ , notice how there is only single value of $\lambda$.
Hence the required plane is , $x-8y+7z+2=0$
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 answers given are shorter, but you could also proceed in the following way: you are given a line (two equations) and a plane (one equation) and in $\mathbb{R}^3$ there are only three possibilities: the line is parallel to the plane and not a part of it, the line intersects the plane or the line is parallel to the plane and a part of it.
Looking at your equations, we can consider them as one system of equations. These systems have either no, one or infinitely many solutions! This corresponds to the previous situations I have described. Using this approach, we find the following system of equations: $$\begin{cases} 2x - y + z &= 1\\ 5x + 3y &= 8\\ x - y +z &= -10 \end{cases}$$ which you can put in an augmented matrix and row reduce. This gives you the following solution: $(x;y;z) = (11; -15,66666 ; -35,66666)$ (where I used $;$ to separate the coordinates and $,$ to denote the decimal parts). Hence we find one solution, and so the line must intersect the plane.
To show you that the line does intersect, I have plotted this in GeoGebra using the 3D option and found an intersection piont $A$ with the coordinates I mentioned: