Your considerations about the quality of the intersection of
two planes are correct. The intersection will be computed
most accurately if the planes are orthogonal. The interection
is not defined for parallel planes and hence it will be inaccurate
for almost parallel planes.
This can be quantified by looking at the normal vectors $n_1, n_2$ of
the 2 planes. The angle $\alpha$ between the planes is the angle
between their normal vectors, so:
$$n_1 \cdot n_2 = |n_1| |n_2| \cos \alpha $$
In the following I assume that the normal vectors are normalized (their length is 1). One formula for the intersection
computes the denominator
$$q = 1 - (n_1 \cdot n_2)^2 = 1 - \cos^2 (\alpha) $$
so this $q$ could be a possible measure of the intersection quality:
if $q < \epsilon$
then the quality is bad. The problem is then ill-conditioned. The best quality is $q=1$.
We get the same measure by considering the direction vector of
the intersection line:
$$u = n_1 \times n_2 $$
It is
$$|u| = |n_1| |n_2| \sin\alpha = \sin \alpha$$
$$|u|^2 = \sin^2 (\alpha) = 1 - \cos^2 (\alpha) = q$$
If $q$ is small then the length of the direction vector is small and the intersection line is poorly defined.
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
If you are in $\mathbb{R}^3$, two lines will generally not intersect. You can find the point that minimizes the sum of the (squares of the) distances to the three lines. Information of how to calculate the distance to the line in three dimensions is at Mathworld, then use a function minimizer. In general three planes will intersect in a point. You can solve the simultaneous equations of the planes to find it. If there are errors in your planes, there will be an error in your point.