Find all planes which are tangent to a surface I'm given the surface $z=1-x^2-y^2$ and must find all planes tangent to the surface and contain the line passing through the points $(1, 0, 2)$ and $(0, 2, 2).$ I know how to calculate tangent planes to the surface given one point, but how would I do so given two points? Any push in the right direction would be appreciated. Thank you.

Given $p_1,p_2$ and $G(p) = 0$ with $p = (x,y,z)$, determine a plane $\Pi\to (p-p_1)\cdot\vec n=0$ with $\{p_1,p_2\}\in \Pi$ and such that $\Pi$ is tangent to $G$ at $p^*$
Representing for $\nabla G(p^*)$ the normal to the the surface $G(p)=0$ at point $p^*$ we have the conditions.
$$
\cases{
\nabla G(p^*) = \lambda\vec n\\
(p_2-p_1)\cdot \vec n = 0\\
G(p^*) = 0\\
(p^*-p_1)\cdot\vec n = 0\\
\|\vec n\| = 1
}
$$
seven equations and seven $(p^*,\vec n,\lambda)$ unknowns.
