Plane equation

Let Q (x, y, z) be a point in the plane.
The line PQ is also in the plane. Hence PQ is normal to the line L: x=-2t, y=t+1, Z=1
Let A be a point on the line L at t = 0,
A = (0, 1, 1)
Let B be a point on the line L at t = 1,
B = (-2, 2, 1)
Therefore,
PQ‧AB = 0
(x-1, y-2, z-3)‧(0-(-2), 1-2, 1-1) = 0
(2)(x-1) + (-1)(y-2) + (0)(z-3) = 0
2x - 2 - y + 2 = 0
2x + y = 0
the equation of the plane : 2x + y = 0

Sorry but this is the whole of the question.

Please state your question carefully