How to find the equation of a plane in the form $ax + by + cz = d$ which contains two lines, $l$ and $m$? The equations of the lines are given:
$l:$ $$r = i + j + k + s(i - j + 2l)$$
$m:$ $$r = 4i + 6j + k + t(2i + 2j + k)$$
We are also told that the lines intersect, and the angle between the two is $74.2°$. That's all the information we have been given.
This is my working:
$$(1,-1,2).(a,b,c) = 0$$ because the dot product would be zero considering the normal of the plane would be perpendicular to the direction vector of both lines.
$$a-b+2c=0$$
$$(2,2,1).(a,b,c)=0$$ using the same principal, but for the line $m$.
$$2a-2b+c$$
The mark scheme says something about solving the ratios and then obtaining the answer, but I did not understand. This is A Level Mathematics P3.
The answer of this problem is $5x-3y-4z=-2$.
I tried finding the cross-product of the two lines, but I'm getting the wrong answer. Even if I do find the cross product, I don't know how to find $d$. 
So how would I solve this and what am I doing wrong? Thanks!
 A: You have two equations in three variables. There are infinitely many solutions, and any one would give you a normal vector. After all, there are infinitely many normal vectors to a plane.
After you get a normal vector, just take an arbitrary point that you know is on the plane and use it to find the remaining constant in the equation for the plane.
A: If I understand correctly $\;i,j,k\;$ is just another name for the coordinate axis, so your lines are
$$\begin{align*}&\ell_1: (1,1,1)+s(1,-1,2)=(s+1,\,-s+1,\,2s+1)\;,\;\;s\in\Bbb R\\{}\\&\ell_2: (4,6,1)+t(2,2,1)=(2t+4,\,2t+6,\,t+1)\;,\;\;t\in\Bbb R\end{align*}$$
Comparing both rightmost expressions, we find the intersection point:
$$\begin{cases}t+1=2s+1\implies t=2s\\2t+6=-s+1\implies s=-2t-5\\2t+4=s+1\implies s=2t+3\end{cases}$$
Lines $\;2-3\;$ give us $\;4t=-8\iff t=-2\;$, and then line $\,1\,$ gives $\;s=-1\;$
so the intersection point is $\;(0,2,-1)\;$, which can be then take as the "anchor" point together with both direction vectors of the lines, and the plane is
$$\pi: (0,2,-1)+s(1,-1,2)+t(2,2,1)$$
For the other form find first the (a) normal to the plane by the vectorial product:
$$(1,-1,2)\times(2,2,1) =\begin{vmatrix}i&j&k\\1&\!\!-1&2\\2&2&1\end{vmatrix}=(-5,3,4)$$
So the plane is $\;-5x+3y+4z+d=0\;$. Inputting $\;(0,2,-1)\;$ gives us $\;d=-2\;$ and thus $\;-5x+3y+4z-2=0\;$ is the plane
