You're right that it's generally impossible to find a plane that contains two given lines. In fact, it's possible only if the two given lines intersect. Fortunately, the two lines $l_1$ and $l_2$ in your problem do intersect; in fact, they intersect at the origin $(0,0,0)$.
To see this, look at the equations. The line $l_1$ has equation $(x,y,z) = t(2,-1,0)$. This says that every point $(x,y,z)$ on the line is some multiple ($t$) of the vector $(2,-1,0)$. In particular, when $t=0$, you get $(x,y,z) = (0,0,0)$. So, the line passes through the origin and has direction $(2,-1,0)$.
If we let $\mathbf{p}_0 = (0,0,0)$, then $l_1$ has equation
$$
(x,y,z) = \mathbf{p}_0 +t(2,-1,0) = (0+2t, 0-t, 0)
$$
which might be the kind of line equation you were expecting.
Similarly, the line $l_2$ passes through the origin and has direction $(0,1,2)$.
The plane containing $l_1$ and $l_2$ has normal vector $\mathbf{n} = (-2, -4,2)$, which you correctly computed as a cross product, and it also passes through the origin. So its equation is $(x,y,z).\mathbf{n} = 0$, i.e. $-2x-4y+2z=0$. The equation $x +2y -z =0$ is simpler, and gives you the same plane.
To get the angle between line $l_3$ and the plane, we can get the angle between the line and the plane normal, and then subtract from 90 degrees.
The line $l_3$ is in the direction of the vector $\mathbf{v} = (2,1,1)$, and the plane normal is $\mathbf{n} = (-2, -4,2)$. The angle $\theta$ between these two vectors is given by
$$
\cos\theta = \frac
{\mathbf{v} \cdot \mathbf{n}}
{|\mathbf{v}| \cdot |\mathbf{n}|}
= \frac
{(2)(-2) + (1)(-4)+(1)(2)}
{\sqrt{2^2 + 1^2 + 1^2}\sqrt{(-2)^2 + (-4)^2 + 2^2}} =
\frac{-6}{12}
$$
Then the angle between the plane and $l_3$ is $90- \theta$.
Best Answer
Take the dot product $n \cdot (a-p)$: if the sign is positive then $a$ exists somewhere past the plane in the direction of the normal $n$. If such dot product is zero then $a$ is in the plane and in the case the dot product is negative is in the opposite side of the plane w.r.t the normal $n$.