Find the line passing thought the point $p=(1,2,0)$, paralel to the plane... Find the line passing thought the point $p=(1,2,0)$, paralel to the plane $P=\{x,y,z \mid x+2y-z=-4\}$ and crossing the line $L=\{(x,y,z):x+2y=2, y+z=4\}$
So I've tried to put the equation of plane and insert the point, because I think that if line is paralel to the plane, then line is contained in the plane.
But I've checked, that the point don't lay in the plane, because
$x+2y-z=1+4-0=5 \neq -4$
So I'm stuck here.
-- EDIT --
Is that correct:
$x+2y-z=k\,\,\, k\in\mathbb{R}$
Intercepting point
$
\begin{cases}
x+2y=z\\
x+2y=2\\
x+z=4
\end{cases}\implies\begin{cases}
z=2\\
x=2\\
y=0
\end{cases}$
General line equation 
$p_{1}-\lambda p_{2}=0 $
$(1-2\lambda,\,\,2,\,\,\,-2\lambda)$
$
x=1-2\lambda$
$y=2$
$z=-2\lambda$
So the answer is 
$x=1-z$
$y=2$ 
 A: It says that the line you need to find is parallel to the given plane, not that it's on that plane.
As this is a homework problem, I'm not going to do it for you, but I will say that you've already done one step, although you may not have realised it.
There are three steps - find plane, find intercept point, construct line through points. I'll leave it to you to figure out what I'm saying from there.
A: Hint: The possible directions of your line are given by the equation
$$
x+2y-z=0
$$

Your solution isn't correct: it is true that $(1,2,0)$ is on your line, but the intersection with $L$ is empty:
$$
\begin{cases}
2+z=4\\
(1-z)+4=2
\end{cases}
$$
gives $z=2$ (first equation) and $z=3$ (second equation.

Here is one possible solution, using Cartesian equations, like yours (my hint would have provided a solution in parametric equations).
What we're going to do is find the required line as the intersection of two planes: one parallel to the given one and the other containing $L$, both passing through $P=(1,2,0)$.
For the first plane we just have to put $(1,2,0)$ in the equation $x+2y-z=k$, giving
$$
\pi_1: x+2y-z=5
$$
Now, the pencil of planes containing $L$ is simply
$$
\mathcal{F}: \lambda(x+2y-2)+\mu(y+z-4)=0
$$
thus to find the plane through $P\notin L$ we just need to substitute $(1,2,0)$
$$
\begin{gather}
\lambda(1+4-2)+\mu(2-4)=3\lambda-2\mu=0\\
3\lambda=2\mu
\end{gather}
$$
therefore our second plane is given by
$$
\pi_2:2(x+2y-2)+3(y+z-4)=2x+7y+3z-16=0
$$
hence the required line is given by
$$
R:\left\{(x,y,z):x+2y-z=5, 2x+7y+3z=16\right\}
$$
You can now check that $R\cap L\neq\varnothing$ and that $P\in R$.
