You are absolutely correct that there will be infinitely-many planes through the line of intersection $U$ of the two given planes. However, given any line $V$ that isn't parallel to $U$, there is only one plane through $U$ parallel to $V$. In particular, the normal vector of that plane is necessarily orthogonal to the direction vectors of both $U$ and $V$. (I discuss this more in this related post.)
For brevity's sake, let's describe the two lines with vector equations. In particular, the line of intersection $U$ is comprised of all points $\langle x,y,z\rangle$ such that $x=7-3y$ and $z=5y-6.$ That is, $U$ can be characterized as the set of all points of the form $$\langle x,y,z\rangle=\langle 7-3s,s,5s-6\rangle=\langle7,0,-6\rangle+s\langle-3,1,5\rangle$$ for some real $s$.
Let $V$ be the other line, so by our above work, if $\langle x,y,z\rangle$ lies on $V,$ then $\frac{x-3}1=\frac{z-2}1$ (so $x=z+1$) and $\frac{y-1}2=\frac{z-2}1$ (which implies that $y=2z-3$). Hence, the points of $V$ are those of the form $$\langle x,y,z\rangle=\langle t+1,2t-3,t\rangle=\langle1,-3,0\rangle+t\langle1,2,1\rangle$$ for some real $t$.
Now, the normal vector to our plane should be orthogonal to both $\langle-3,1,5\rangle$ and $\langle1,2,1\rangle,$ so a convenient choice is the cross-product $$\langle a,b,c\rangle=\langle-3,1,5\rangle\times\langle1,2,1\rangle=\langle-9,8,-7\rangle.$$ Now we can choose any point $\langle x',y',z'\rangle$ on $U$--for simplicity, say $\langle x',y',z'\rangle=\langle7,0,-6\rangle$--and we have our plane equation $$-9(x-7)+8(y-0)-7(z+6)=0.$$
You just need one vector $n$ that is orthogonal to both lines (you can take the cross product of the direction vectors of the lines as you did). Then if you take a point $p_1$ on the first line the plane is given by all points $v$ so that $(v-p_1)\cdot n=0$ where $\cdot$ is the dot product. You can do the same thing for the second plane (use the same normal vector $n$).
If you expand this out, you should get that the plane is given by
\begin{align*}0&=((x,y,z)-(1,2,3))\cdot(3,1,5) \\ &=(x-1,y-2,z-3)\cdot(3,1,5) \\ &=3x-3+y-2+5z-15 \\ &=3x+y+5z-20\end{align*}
as the equation for the first plane. Just do the exact same working for the second plane, by subtitute $(1,-1,1)$ for $(1,2,3)$.
Best Answer
The line passes $z=0$ at $(1.2, 2.8, 0)$.
A vector from this point to $(-3,2,1)$ is $(4.2,0.8,-1)$.
This vector cross $(4,-9,-5)$ is $(13,-17,41)$, which is the normal of the plane.
The result is:
$$13(x-(-3))+(-17)(y-2)+(41)(z-1)=0$$
This is the scalar product of a vector in the plane and the normal vector.