Let P = (3, 2, -1) , Q = (1, -1, 2) and R = <3, 2, -1> = direction vector from r(t)Find the equation of the plane passing through the points (3, 2, -1) and (1, -1, 2) that is parallel to the line r(t) = <1, -1, 0> + t<3, 2, -1>

a = PQ = <1, -1, 2> - <3, 2, -1> = <-2, 3, -3>

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?