Equation of plane 
Find the equation of the plane through the point $(1,−1,2)$ which is
  perpendicular to the curve of intersection of the two surfaces
  $x^2+y^2−z=0$ and $2x^2+3y^2+z^2−9=0$.

i've gotten as far as subbing one equation into the other but i'm stuck on the differentiation. it would be much appreciated if someone could help with this

Outline: If we substitute in this way:
$$2\overbrace{(z-y^2)}^{x^2}+3y^2+z^2-9=0$$
and 
$$2x^2+3\overbrace{(z-x^2)}^{y^2}+z^2-9=0$$
we can obtain equations for $x$ and $y$ in terms of $z$.  So the curve through the point $(1,-1,2)$ can be parameterized by $$\big(\sqrt{\text{function of } t},-\sqrt{\text{function of } t},t\big)$$ and the derivative of this vector when $t=2$ (i.e., $z=2$) will be normal to the plane.  Given a normal vector, and a point $(1,-1,2)$ on the plane, we can determine the equation of the plane.
