a. Find a point at where these lines intersect
b. Find the equation of a plane that contains the two lines.
r[/B] = <1,3,0> + t<3,-3,2>
r = <4,0,2> + s<-3,3,0>
The Attempt at a Solution
I correctly found the point of intersection to be (4,0,2) for part a, but for some reason, I can't find the equation of the plane. I solved the cross product for <3,-3,2> and <-3,3,0> to get <-6,-6,18> and used that to find -6(x – 4) - 6y + 18(z – 2) = 0 as my equation, but that was marked incorrect. Is this an arithmetic issue or did I approach the problem the wrong way? I've gone over it numerous times and can't find the error.