What Plane contains the line? I have got this line
$$\mathrm g:\;\overrightarrow{\mathrm x\;}\;=\;\begin{pmatrix}-1\\2\\0\end{pmatrix}\;+\;\mathrm s\cdot\begin{pmatrix}-1\\4\\-1\end{pmatrix}$$
And this plane: 
$$E_t:\,\, tx + y + tz = 0$$
Now for which t is the line element of the plane? 
My calculations:
When I put the equation of the line in the equation of the plane i get:
And this plane: 
$$ t(-1-s) + (2+4s) + t(-2) = 0$$
This is the same as:
$$2-t+s(4-2t)=0$$
$$=> s=\tfrac{t-2}{4-2t}=-\tfrac{1}{2}$$
And now I do not know how to go on.
The solution is $t=2$
Can someone explain the last step to the solution please?
Thanks
 A: The line and the plane are parallel iff the plane's normal vector $(t,1,t)$ and the line's direction vector $(-1,4,-1)$ are perpendicular; this gives $t=2$.
As for $t=2$ the vector $(-1,2,0)$ lies in the plane the whole line lies in that plane.
A: It seems like your calculation has some error by plugging in the equation of the line in the one of the plane, as noticed by José Carlos Santos..
For another solution: it is necessary for the line to be included in the plane that $(1,-6,2)\in E_t  \iff t-6+2t=0 \iff t=2$.
This is also sufficient, since we can now check that $2(1-s)+(-6+4s)+2(2-s)=0$ for all $s\in \mathbf{R}$.
A: Calling 
$$
p = (x,y,z)^{\dagger}\\
\vec n = (t,1,t)^{\dagger}\\
p_0 = (-1,2,0)^{\dagger}\\
\vec v = (-1,4,-1)^{\dagger}
$$
we have the plane $\Pi\to p\cdot \vec n=0$ and the line $L\to p=p_0+s\vec v$. Now if $L\in \Pi$ we have
$$
(p_0+s\vec v)\cdot \vec n = 0\Rightarrow\cases{p_0\cdot\vec n = 0\\
\vec v\cdot\vec n = 0}
$$
From the second equation we have $-t+4-t = 0\Rightarrow t = 2$ and now making $p_0\cdot \vec n= 2\times (-1)+1\times 2+ 2\times 0 = 0$ hence for $t=2\Rightarrow L\in \Pi$
