The general equation of a plane in 3-D is given by $$\mathbf{(p-p_0).n}=0$$
where $\mathbf{p}$ is any general point on the plane, and $\mathbf{p_0}$ is any known point on the plane. $\mathbf{n}$ is a vector normal to the plane.
The equation of a line is given by $$\mathbf{p = l_0+}t\mathbf{l} $$
where $\mathbf{l_0}$ is any point on the line.
If the line lies in the plane, it must satisfy two conditions-
It must be perpendicular to the normal to the plane i.e. $\mathbf{l.n}=0$
$\mathbf{l_0}$ must lie in the plane i.e. satisfy the plane's equation. So, $\mathbf{(l_0-p_0).n} = 0$
You can calculate $\mathbf{p_0,l_0}$ very easily with the information you have.
$\mathbf{p_0}$ can be calculated by choosing any two of $x,y,z$ and finding the third to satisfy the equation of the plane. $\mathbf{l_0}$ is precisely $\vec O$ that you already know. You can use them to cross-check whether the fit is good or not. But they do not depend on $\mathbf{l}$. So, they are secondary, and may be used as a sanity check later on.
Suppose you are given the equation of the line as $ax+by+cx+d = 0$. You can recast it as $[(x,y,z)-\mathbf{p_0}].(a,b,c)$. So, $(a,b,c)$ is your normal vector.
Suppose you have $m$ planes with normals $\mathbf{n_1,n_2,\ldots,n_m}$. Then, your overall constraints are $$\mathbf{l.n_1} = 0 \\ \mathbf{l.n_2} = 0 \\ \ldots \\\mathbf{l.n_m}=0$$.
It is a system of linear equations with 3 variables and $m$ equations. You need to find out $\mathbf{l}$, which can be found up to a constant factor by the standard least squares method, provided $m>3$, which should not be a worry for you. If you code carefully enough, you can implement all of it in matrix terms.
Hope it helps.
The key thing you pointed out is that $Ax = b$ having a unique solution for some $b$ shows that the columns of $A$ are linearly independent. In particular, the columns of $A$ are three linearly independent vectors in $\mathbb{R}^3$, so they actually form a full basis for $\mathbb{R}^3$. By the definition of a basis, every vector in $\mathbb{R}^3$, such as the vector $(2,3,4)$, can be uniquely written as a linear combination of $A$'s columns. This unique linear combination corresponds to the unique solution to $Ax = (2,3,4)$, so the answer is (B).
Best Answer
For the first part of your question, adding the two planes does not yield their line of intersection. In fact, it does not even yield a line, it is the equation of a plane passing through their line of intersection.
In general, given two planes $P_1,P_2$, $$P_1+\lambda P_2=0, \lambda\in\mathbb{R}$$ represents the family of planes passing through the line of intersection of $P_1$ and $P_2$.
For the second part, taking the cross product would yield the direction ratios of the line. To find a point on the line easily generally one of $x,y$ or $z$ are set equal to zero, and the two equations are solved to get a point.
As you noticed, it is possible that the line does not intersect the plane $z=0$ in which case you can easily observe that the set of equations will be inconsistent.
In that case, put one of the other two as equal to zero. Obviously the line has to intersect two of the three planes $\{x=0,y=0,z=0\}$.