 Tangent plane requires the point to lie on the surface

For a level set $G=0$, the tangent plane formula $\nabla G(p)\cdot (x-p)=0$ requires both that $p$ belongs to the surface and that its gradient is nonzero. Evaluating the gradient at an off-surface point instead gives a plane tangent to the level set through that different point. For example $G=x^{2}y-4z{e}^{x+y}+11$ has $G(2,-2,9)=-33$, so the gradient plane there cannot be a tangent plane to $G=0$. Holding the first two coordinates fixed gives the on-surface point $(2,-2,3/4)$ and its plane $11x-y+4z=27$.