This isn't "PR". It would be RQ, which is the negative of QR, except that you have subtracted wrong. For the second component, -1- 2= -3, not 3 and for the third component 2- (-1)= 3, not -3.

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 = 1So 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.