= Tangent plane requires the point to lie on the surface
{title2=$G(p)=0,\quad\nabla G(p)\ne0$}
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 $G=x^2y-4ze^{x+y}+11$ 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 $11x-y+4z=27$.
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