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 $$-9(x-7)+8(y-0)-7(z+6)=0.$$
In reply to comments:
Well, there's a few different ways to define a plane. There's the way you will know which is a plane is defined by a point and two vectors in the plane $\mathbf r = \mathbf a + \lambda \mathbf b + \mu \mathbf c$. That's all you need to express everything we need to know about the plane in order to define it uniquely, but there's a way of simplifying this using the cross-product (if you haven't done the cross product, its an operation which combines two vectors to generate a third vector that is perpendicular to both).
If we take the cross product of the two lines in the plane, we get a (nearly) unique vector that's perpendicular to the plane. I say nearly because the vector could point the other way. Anyway, since this vector is perpendicular to the plane, we can say that the dot product of our vector with any vector in the plane is zero. So this gives us another way of expressing a plane. If $\mathbf r$ is the position vector of a general point in the plane, $\mathbf a$ is the position vector of a specific point in the plane, and $\mathbf v$ is a vector perpendicular to the plane, then:
$$ (\mathbf r - \mathbf a)\cdot \mathbf v = 0$$
We can link this back to our original equation by noting that $\mathbf v = \mathbf b \times \mathbf c$ is a vector perpendicular to the plane.
There's one final way of defining a plane - in Cartesian coordinates. It doesn't involve vectors though, and you'll probably come across it soon if you haven't already, so I won't explain it.
Best Answer
You are right in taking the required plane to be $P_1+k P_2 = 0$ (Intersection of $P_1$ and $P_2$). Since the required plane is parallel to the $x$-axis, the co-efficient of $x$ in it's equation should be $0$. $2k+1 = 0 \implies k = \frac{-1}{2}$. The required plane is: $$ \frac{-y}{2} + \frac{3z}{2} - 3 = 0 $$
BTW, the $z$ co-efficient is $1-k$ and not $1-4k$.