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

$p_{1}-\lambda p_{2}=0$

$(1-2\lambda,\,\,2,\,\,\,-2\lambda)$

$x=1-2\lambda$

$y=2$

$z=-2\lambda$

$x=1-z$

$y=2$

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You find it.${}{}{}{}$ – Alexander Gruber Mar 22 '13 at 9:52
Please have a thorough look at the FAQ about homework questions. – A.P. Mar 22 '13 at 10:02
OK, I'll look now – Joggi Mar 22 '13 at 10:04
@AlexanderGruber Fixed. I don't know how to find it – Joggi Mar 22 '13 at 10:07

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.

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How to find intercept points? I need to solve equation system with the $x+y2-z=k$ and equations for line $L$ ? – Joggi Mar 22 '13 at 10:33
This point will be (2,0,2) ? – Joggi Mar 22 '13 at 10:36
You need to find the plane first. You haven't done that, yet. – Glen O Mar 22 '13 at 11:03
If at any point you think you've solved it, check if it satisfies the three conditions - does it pass through the appropriate point? Does it have an intercept with the given line? Is it parallel to the given plane? – Glen O Mar 22 '13 at 11:09

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 $P\in R$.

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So every paralel plane is described by $x+2y-z=0$? – Joggi Mar 22 '13 at 10:23
No, every parallel plane is described by $x+2y-z=k$ with $k\in \Bbb R$ (same orthogonal direction, but passing through different points). If you are taking values in $\Bbb R$, of course. – A.P. Mar 22 '13 at 10:24
Thanks. Could You check if I've done it correctly, in question edit? – Joggi Mar 22 '13 at 10:43
I'm sorry but no, it isn't. – A.P. Mar 22 '13 at 10:56
I provided a full solution in my edit. – A.P. Mar 22 '13 at 11:20