Find two points on the line. Take the vectors from those two points to A, and take the cross product. That'll be your normal vector. Use point A, or one of the other points you found as a point in the plane.
You do have two equations:
and
To find two different points, simply pick an x-coordinate, and then solve for y and z.
I have not seen two different points on the line, only one: (-1,2,0). You need another point on the line, because you have to take the cross product of two linearly independent vectors in the plane. (Incidentally, it would probably impress your teacher if you showed that point A was not on the line.)
You were given the line as
That's NOT one equation, it is two : and
(strictly speaking, there is a third equaton, but that's obviously not independent of the first two.)
Take x= 6 (so that is easy) and then you have so that y= 6 and so that z= -1. One point on the line is (6, 6, -1).
Now take x= -8 so that . Then so that y= -2 and so that z= 1. Another point on the line is (-8, -2, 1).
Can you find the equation of the plane that contains the points (6, 6, -1), (-8, -2, 1), and (1, 3, 2)?