Question

Calc 3..Find an xyz equation for the plane that is perpendicular to {2,-1,-1} and goes through the point {1,-1,0}

Answer 1

(2,-1,-1) is the normal vector? or just a point?

Answer 2

The equation of the plane is
2 x -y -z = d\\
to find d. The plane has to go through the point (1,-1,0}
2(1) -(-1) -0 =d
2 + 1 =d
d=3
The equation of the plane is
2 x -y -z =3

Answer 3

seems right except that that would be containing the given vector, as opposed to being orthogonal.

Answer 4

and yes sorry that is the normal vector