finding an equation of a plane Given the lines:
$l_1: (1,-2,5)+t(2,1,-1)$
$l_2: (3,4,1)+s(2,1,-1)$.
Need to find the equation of the plane that is equally distance from those two lines, and Perpendicular to a plane that those lines creates.
My attempt: 
I can see that the lines are parallels. i tried to find the equation of the plane that the two lines form but i don't know how to continue.
 A: Hint the equation of plane containing two lines is $(r-a_1). (b_1×b_2) $ where $a_1$ can be any of the two points. And $b_1,b_2$ are two vectors. Now find the point on the required plane which is at same distance from both lines. This can be done like $x-1=|x-3|$ same with $y,z $ to get the point as $-1,-3,3$ now this plane is perpendicular to other plane containing the lines which means the normals are perpendicular so let drs of normal of unknown plane be $l,m,n $ we can get drs of normal of other plane ie plane containing lines from the above mentioned equation to get $ll_1+mm_1+nn_1=0$ get ratios in terms of $l$  of $l,m,n $ . Equation of a plane is also $lx+my+nz=d $ where $l,m,n $ are drs and $d $ is a constant . This plane passes through $(-1,-3,3) $ to get the required plane.
A: Let's rewrite the two lines as follows
$$
\begin{gathered}
  l_{\,1} :\mathbf{a} + t\,\mathbf{v} = \left( {1, - 2,5} \right) + t\left( {2,1, - 1} \right) \hfill \\
  l_{\,2} :\mathbf{b} + t\,\mathbf{v} = \left( {3,4,1} \right) + s\left( {2,1, - 1} \right) \hfill \\ 
\end{gathered} 
$$
so that it is clear the definition of the two points $\mathbf a$ and $\mathbf b$ and vector $\mathbf v$.
Then as you correctly noted, being the vector $\mathbf v$ the same, the lines are parallel.
Consider now the segment joining the points  $\mathbf a$ and $\mathbf b$ and the relevant difference vector
$$
\mathbf{c} = \mathbf{b} - \mathbf{a} = \left( {2,6, - 4} \right) = 2\left( {1,3, - 2} \right)
$$
The plane $\pi _{\,//} $ containing the two lines, shall contain both points and be parallel to $\mathbf v$,
i.e. it shall pass by one of the points and be normal to the vector
$$
\mathbf{n} = \mathbf{c} \times \mathbf{v} = \left( { - 2, - 2, - 4} \right) =  - 2\left( {1,1,2} \right)
$$
and since the vector $\mathbf n$ is not null, the two lines are distinct.  
Now the required plane $\pi _{\, \bot } $ shall clearly be:
 - orthogonal to  $\pi _{\,//} $, i.e. parallel to $\mathbf n$
 - parallel to the lines, i.e. parallel to $\mathbf v$
 - equidistant from the lines, i.e. passing through the mid-point of the segment $\mathbf b - \mathbf a$  
which means:
 - orthogonal to the vector $\mathbf{m} = \mathbf{n} \times \mathbf{v} = \left( { - 3,5, - 1} \right)$
 - passing through the point $\left( {\mathbf{b} + \mathbf{a}} \right)/2 = \left( {2,1,3} \right)$  
So we get
$$
\pi _{\, \bot } :\left( {x - 2,y - 1,z - 3} \right) \cdot \left( { - 3,5, - 1} \right) = 0
$$
