Point normal equation of plane Im doing an homeassignment in linear algebra!
This is the question I'm having problems with!

The plane M contains two lines;
l1 : (x, y, z) = t(2, −1, 0), t ∈ R,
l2 : (x, y, z) = t(0, 1, 2), t ∈ R.
Determine a point-normal equation of the plane M and compute the acute angle between M and the line;
l3 : (x, y, z) = t(2, 1, 1), t ∈ R. (ON-system assumed.)

I have taken the crossproduct of l1 and l2 and got that computed to be: (-2, -4, 2) , which in my understanding is n (the normal of the plane?) But them I am stuck.I can't understand if the plane contains fully the two lines, or only points from the lines?? From what i've learned two lines are too much information for a plane in 3-dim? Three points or one line and one point is enough.?? Also why is the lines only described in parametric (x, y, x) = t(0,1,2). What does the parameter exactly mean, is it a direction vector? usually is lines not described with: {x-3 = t, y+6 = 3t, z+5 = 2t}...I am probably not understanding the concept completely, I have been trying to do this question for hours now and I really need some help to get me going. 
I also know that for shortest distance I want the angle of line and plane to be pi/2, since then cos(theta) will be 0. I can probably also use this somehow??
Grateful if I could get some concrete tips and tricks how to do this and what it all really means...
From where do I get the point p0 and p, are they from the lines in the plane or from the line i will compute the acute angle of l-M??
Have I calculated n (normal) to plane correctly??
Why are the lines only described in parameter t??
 A: 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 $90- \theta$. 
A: Take the cross product of l1 and l2 directions to find the plane normal vector (non-unit vector)
$$ \vec{n} = \vec{e}_1 \times \vec{e}_2 = (2,-1,0) \times (0,1,2) = (-2,-4,2) $$
The point-normal equation for a plane is $$\vec{n} \cdot (\vec{r}-\vec{r}_0) =0$$ which works of any scalar multiple of $\vec{n}$ (no need to normalize to a unit vector) and $\vec{r}_0$ as any point on the plane. Might as well take $t=0$ for l1 to give $\vec{r}_0 = (0,0,0)$. So the plane equation is
$$ -2x-4y+2z=0 $$
The acute angle $\theta$ of the plane to the line l3 is found by noting that it is equal to 90° minus the angle $\varphi$ between the line and the plane normal. This is found by the inner product rule
$$ \vec{n} \cdot \vec{e} = | \vec{n}| |\vec{e}| \cos\varphi $$
with $\vec{n}=(-2,-4,2)$ and $\vec{e}=(2,1,1)$
$$\varphi = \cos^{-1} \left( \frac{\vec{n}\cdot\vec{e}}{|\vec{n}| |\vec{e}|} \right) = \left( \frac{(-2,-4,2)\cdot(2,1,1)}{ (2\sqrt{6})(\sqrt{6})} \right) = \cos^{-1} \left( \frac{-6}{12} \right) = 120°$$
Since the angle is > 90° we have to flip the direction of the line to get $\vec{e} = (-2,-1,-1)$ and 
$$ \varphi = \cos^{-1} \left( \frac{6}{12} \right) = 30° $$
So the angle of the line l3 to the plane is $$\boxed{\theta = 90° - \varphi = 60°}$$
