Are you sure its right cos i tested it at this online solution thing and it said it wasnt:
Hint: the equation of a plane, in general, is given by expanding
r •n = r0 •n
where r = [x, y, z], r0 is the position vector of a point on the plane, and n is a normal to the plane.
Now, rearrange the given equation in the form f(x, y, z) = 0. Then ∇f has the property of being normal to the surface at any point on the surface defined by f(x, y, z) = 0. Thus ∇f evaluated at (2, 0, 0) and (2, 0, 0) itself, are respectively, a normal and a point on the tangent plane you seek.