Finding a plane parallel to and equidistant from two lines So i have to write down the equation of a plane that is parallel to this two lines:
$$p1: y=2x-1 , z=3x+2 $$
$$p2: y=-x+1, z=4x-1$$
And the plane is at the same distance from both of those lines.
I was thinking i should that the dot product of normal vector and each of those lines direction vectors should equal 0, but i really don't know how to continue, beacause i get a not so simple equation and i still have two variables when i simplify it.
Also i don't how to use information that the planes is at the same distance away from line 1 as from line 2. I would really appreciate any help in solving this problem. 
 A: Here is one way:
Line $p_1$ is defined by $(1,2,3)x+(0,-1,2)$, while $p_2$ is defined by $(1,-1,4)x+(0,1,-1)$.
So the direction vector for $p1$ is $(1,2,3)$ and the direction vector for $p2$ is $(1,-1,4)$. A quick way to find a vector perpendicular to both of them is to calculate their cross product, which is $(11,-1,-3)$.
Therefore any plane that is perpendicular to the vector $(11,-1,-3)$ will be parallel to or coincident with your two given lines. The equation of such a plane is
$$11x-y-3z=c$$
for some constant $c$.
If the plane is equidistant from the two lines, then the value of $c$ will be equidistant from the values of that expression when a point on each line is substituted. The value of $11x-y-3z$ for point $(0,-1,2)$ on line $p_1$ is $-5$, while the value for point $(0,1,-1)$ on line $p_2$ is $2$.
Therefore we want $c$ to be the average of $-5$ and $2$, which is $-\frac 32$. Your desired plane is
$$11x-y-3z=-\frac 32$$
A: Equation of p1 & p2 are :$$l1 : (0,-1,2)+t(1,2,3)$$$$l2 : (0,1,-1)+t(1,-1,4)$$
Notice that they don't intersect and thus we can find the required plane.
To find the direction $v : (v1,v2,v3)$ perpendicular to p1 & p2, solve  $(v1,v2,v3).(1,2,3)=0$ 
$(v1,v2,v3).(1,-1,4)=0$
solution : $t(-11,1,3)$ 
Now lets write the equation of our plane with the above normal : $$p:-11x+y+3z+d=0$$
To find $d$ equate distance of any point from l1 to p with any point from l2 to p :
Recall that the distance of the point$(x_0,y_0,z_0$) from the plane $ax+by+cz+d$ is given by $\frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}} $
 so equating distance of $(0,-1,2)$ & $(0,1,-1)$ from p, we get
$$|5+d|=|-2+d|$$ $$d=-3/2$$
Thus our required plane is $$p:-11x+y+3z-3/2=0$$
