Equation of a plane (not sure if I got this) When I was doing my calculus midterm, I came across a question that I didn't really know how to solve, I think I skipped over these problems in my studies. The question is:
Find an equation of the plane through $(1,2,-2)$ that contains the line:
$x=2t$   $y=3-t$   $z=1+3t$
Since I didn't know how to solve it the "correct" way, what I tried doing was finding two points on the line by plugging in $t=0$ and $t=1$ and I got the points $(0,3,1)$ and $(2,2,2)$.
So now I needed to find the equation of a plane containing $A(1,2,-2),B(0,3,1),C(2,2,4)$.
I found AB and AC and took the cross product and then just plugged in the values into: 
$a(x_0-x)+b(y_0-y)+c(z_0-z) = 0$ 
Does this make sense? Did I salvage at least some partial credit for this answer? Thanks

If $(a, b, c)$ are the components of your normal vector (from the cross product), and $(x_0, y_0, z_0)$ is one of the points in the plane, that looks good!
The way I've seen it before was with terms like $(x-x_0)$ instead of $(x_0 - x)$ but that shouldn't make a difference.  If $\vec N$ is a normal vector, so is $-\vec N$.
