I have to find the eq. of the plane which contains the following lines:

x = 1 - t

y = 2 + 3t

z = 2t

and

x = 1 - s

y = 5

z = 4 - 2s

So we have the vectors <-1, 3, 2> and <-1, 0, -2>

I found the cross product which is <-6, -4, 3> which is my normal vector.

Now what?

I know a_1(x - x_1) + a_2(x - x_2) + a_3(x - x_3) = 0; I have my a_1, ..., a_3, but how do I get the points? Apparently the point is (1, 2, 0), but I don't see how to get that.