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.
I think I can clear up some misunderstanding. A line contains more than just two points. A line is made up of infinitely many points. It is however true that a line is determined by 2 points, namely just extend the line segment connecting those two points.
Similarly a plane is determined by 3 non-co-linear points. In this case the three points are a point from each line and the point of intersection. We are not creating a new point when the lines intersect, the point was already there.
This is not the same thing as saying that there are 5 points because there are two from each line and the point from their intersection.
Best Answer
The equation $ax+by+cz+d=0$ defines a plane in the sense that $$P_1=\{(x,y,z)\in \mathbb{R}^3: ax+by+cz+d=0\}$$ is considered a plane (for good reason). One the other hand, the $(x,y)-$plane $P_2$ is defined by the equation $z=0$. So, to find the intersection $P_1\cap P_2$, you should find the set of points satisfying both sets of equations. That is, you should find the solution set in $\mathbb{R}^3$ of the system $$ \begin{cases} ax+by+cz+d=0\\ z=0. \end{cases}$$ Unsurprisingly, it is given by simply setting $z=0$ in the first equation: $$ P_1\cap P_2=\{(x,y,0): ax+by+d=0\}.$$ Notice that this is a line in the plane $P_2$.