In $\mathbb R^3$, consider the reference planes $x = 0, x = 1, y = 0, y = 1, z = 0, z = 1$, and one line in each plane: $\ell_{x=0}, \ell_{x=1}, \ell_{y=0}, \ell_{y=1}, \ell_{z=0}, \ell_{z=1}$.
For which set of $\{\ell_n\}$ is there a plane $P$ that includes all six $l_n$?
My guess is below, though I've not proven it:
Conjecture: There exists such a plane $P$ iff there exist $A, B, C, D \in \mathbb R, \neq 0$ such that the $\ell_n$ include these points:
$$\begin{align*} \ell_{x=0}:& (0, 0, D/C) \quad (0, D/B, 0)\\ \ell_{x=1}:& (1, 0, (D-A)/C) \quad (1, (D-A)/B, 0)\\ \ell_{y=0}:& (D/A, 0, 0) \quad (0, 0, D/C)\\ \ell_{y=1}:& ((D-B)/A, 1, 0) \quad (0, 1, (D-B)/C)\\ \ell_{z=0}:& (D/A, 0, 0) \quad (0, D/B, 0)\\ \ell_{z=1}:& ((D-C)/A, 0, 1) \quad (0, (D-C)/B, 1). \end{align*}$$
Furthermore, we can assume $D=1$.
Let $\{ \ell_n\}$ as in the OP and assume at least one of the following conditions hold
$\ell_{x=0}$ and $\ell_{y=0}$ are coplanar, not parallel and at least one of them is not parallel to a coordinate axys
$\ell_{x=0}$ and $\ell_{y=0}$ are parallel, distinct and both of them do not coincide with the $z$-axys
$\ell_{x=0}$ and $\ell_{y=0}$ coincide i.e. they coincide with the $z$-axys and $\ell_{z=0}$ is not parallel to a coordinate axys
Than $\{\ell_n\}$ are coplanar lines. If $\{\ell_n\}$ do not satisfy any of the condition than does not exists a plane $P$ containing all six the $\ell_n$-s
Proof
Two distinct lines on $\mathbb{R}^3$ uniquely determine a plane.
As a line in $\mathbb{R}^3$ is defined as the intersection of two not parallel planes and we know that $\ell_n$ is contained in one of the reference planes if $\ell_{x=0}$ does not coincide with $\ell_{y=0}$ and they are coplanar it is enough to find the plane identified by this two lines and impose that such plane is not parallel to any of the reference planes.
So let $\ell_{x=0}$ an arbitrary line with direction $(0,a_1,b_1)$ and passing throught the point $P_{x=0}=(0,y_1,z_1)$ and similarly $\ell_{y=0}$ has direction $(a_2,0,b_2)$ and goes throught the point $P_{y=0}=(x_2,0,z_2)$.
If $\ell_{x=0}$ is not parallel to $\ell_{y=0}$
Consider the vector $P_{x=0}-P_{y=0}$ the two lines are coplanar iif the triple scalar product is zero i.e $$ -x_2a_1 b_2+ y_1 b_1 a_2 - a_1 a_2(z_1-z_2)=0 $$ If this condition is satisfied the equation of the plane containing the two lines is $$ a_1 b_2 x + b_1 a_2 (y-y_1)-a_1 a_2(z+z_1)=0 $$ If at least one between $b_1$ and $b_2$ is not zero (i.e. at least one of $\ell_{x=0}$ or $\ell_{y=0}$ is not parallel to a coordinate axys), this plane is not parallel to any of the reference planes and so it characterizes the remaining lines. e.g. $$ \ell_{x=1} : \begin{cases}a_1 b_2 x + b_1 a_2 (y-y_1)-a_1 a_2(z+z_1)=0 \\ x=1 \end{cases} $$
Otherwise it is parallel to the $z=0$ plane and so it is impossible for the $6$ lines to be coplanar
If $\ell_{x=0}$ is parallel to $\ell_{y=0}$
if they are distinct the two lines identify the plane $$ y_1 (x-x_2)+ x_2 y =0 $$ If $y_1 \ne 0 \ne x_2$ (i.e. $\ell_{x=0}$ and $\ell_{y=0}$ are not the $z$-axys) the plane is not parallel to any of the reference planes and so this characterizes all the remaining lines. Otherwise the plane is parallel to $x=0$ or $y=0$ and so it is not possible for the $6$ lines to be coplanar
If they coincide they coincide with the $z$-axys. Consider now an arbitrary line $\ell_{z=0}$ with direction ratio $(a_3,b_3,0)$. If $(0,0,0) \notin \ell_{z=0}$ the line is not coplanar with the $z$-axys. Otherwise the equation for the plane containing the $z$-axys and $\ell_{z=0}$ is
$$ b_3x - a_3 y=0 $$
If $a_3 \ne 0 \ne b_3 $ (i.e. $\ell_{x=1}$ is not parallel to a coordinate axys) this plane is not parallel to any of the reference planes i.e. it characterizes all the remaining lines e.g. $$ \ell_{x=1} : \begin{cases} b_3 x- a_3 y=0 \\ x=1\\ \end{cases} $$
(this disproves your conjecture as it is possible to let $\ell_{x=0}$ be the $z$-axys. Thus $\left(0,\frac{D}{B},0\right) \notin \ell_{x=0}$ unless $D=0$)
Otherwise the plane is parallel to either $y=0$ or $x=0$ and, as before it is impossible that all the lines are coplanar