How do I find a plane that contains two given lines?

[Mauve]

Homework Statement

a. Find a point at where these lines intersect
b. Find the equation of a plane that contains the two lines.

Homework Equations

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.

 

[Orodruin]

Double check your cross product.

 

[Mauve]

Double check your cross product.

So apparently 9 - 9 = 0 and not 18 and also I am a raging imbecile. Thank you very much for your input. It really did help

 

[Orodruin]

So apparently 9 - 9 = 0 and not 18 and also I am a raging imbecile.

Arithmetic error happens to everyone and when your brain has told you one thing once it may be difficult to find it until someone points it out.