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 $$-9(x-7)+8(y-0)-7(z+6)=0.$$
Best Answer
Let's write the equations of the planes as
$$z=3x+2y-28 \tag 1$$
and
$$z=-\frac12 x+2y \tag 2$$
If the planes intersect, then obviously the values of $z$ on the line of intersection must be equal. So, setting the right-hand sides of $(1)$ and $(2)$ equal yields
$3x+2y-28=-\frac12 x+2y\implies x=8$
Thus, the two planes intersect at a line on the plane $x=8$. To find the line in the plane $x=8$, we need only look on either plane and see the relationship between $y$ and $z$. To that end, letting $x=8$ in $(1)$ reveals that
$$z=2y-4$$
Therefore, the line of intersection is given parametrically by
$$\vec r(t)=(\hat x8+\hat y2)+(\hat y +\hat 2z)t$$