Yes. Repeat what you did but this time do it correctly.b = QR = <3, 2, -1> - <1, -1, 2> = <2, 3, -3>
N = a x b = <0, -12, -12>
Now I'm going to choose a point on the plane, P = (3, 2, -1)
d = n dot OP = <0, -12, -12> dot <3, 2, -1> = -12(2) + 12 = -12
So it follows that the plane has equation -12y + -12z = -12 or y + z = 1 So it appears that using the direction vector, as I did, from r(t) and generating the plane using those three points is completely wrong. Any guidance?
You don't actually need to find "PR" or "QR" (technically, you can't- R is a vector not a point). Just use PQ, a vector in the surface, and R. R is not necessarily in the surface but it is parallel to it so the cross product of PQ and R will be perpendicular to the surface.