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.
Generally when people talk about planes in three dimensions, they mean a set of the form $P=\{ x \in \mathbb{R}^3 | n^T x = d\}$, where $n \in \mathbb{R}^3$ is a fixed non-zero vector and $d \in \mathbb{R}$.
A plane is infinite in extent as $\ker n^T$ is two dimensional, and the point $x_0={d \over \|n\|^2} n \in P$, so $x_0+k \in P$ for any $k \in \ker n^T$.
If the plane passes through the origin, it is easy to see that the corresponding $d$ above must be zero.
The intersection of two planes that pass through the origin will be given by $L = P_1 \cap P_2$, where $P_k = \{ x \in \mathbb{R}^3 | n_k^T x = 0\}$.
We can write
$L = \{ x \in \mathbb{R}^3 | n_1^T x = 0, n_2^T x = 0\} = \ker N^T$, where $N = \begin{bmatrix} n_1 & n_2 \end{bmatrix}$.
If the planes are different (equivalently $n_1, n_2$ do not lie on the same line), then $n_1, n_2$ are linearly independent, and $\dim \ker N^T = 1$, hence $L$ is a line of infinite extent passing through the origin.
You can always choose subsets $S_1, S_2$ of $P_1, P_2$ so that $S_1$ and $S_2$ only intersect at the origin, but then $S_1,S_2$ will not be planes in the usual sense.
Best Answer
edit, tl;dr: What usually is meant by two planes being orthogonal to one another in geometry is their normals being orthogonal to each other.
In other words: two one-dimensional subspaces being orthogonal to each other.
2 planes have their normals being orthogonal to each others are sometimes said to be orthogonal. There it is a specific geometric orthogonality pointing out that the normals of the planes are orthogonal to each other. When talking about subspaces orthogonal to each other what is usually meant is all their vectors are pairwise orthogonal. But you can verify for yourself that 2 2D subspaces can not have 0 vector intersection in $\mathbb R^3$.
But if you think about it closer, you will see that the geometric meaning of normals being orthogonal to each other actually means the complement of set 1 and the complement of set 2 are orthogonal to each other. So there is a connection to the same orthogonality concept, but the subspaces are 1 dimensional.