In a given problem, the goal is to find a tangent plane of an equation $x² + 2y² +3z² = 6$ that also includes some points $(6,0,0)$ and $(0,3,0)$. The points should not be relevant for the question I'm asking however.
The answer to the problem is a plane $x + 2y + 3z = 6$, but is this equation really a tangent to the plane? If we for example set $y=0$ and $z=0$, then $x=\sqrt{6}$ in the original equation but $x=6$ in the plane equation and thus cannot include the same tangent point. Doesn't that mean that the plane is not a tangent plane to the surface, or doesn't a tangent plane need to include the point that it is tangent to?
I'm comparing this to having a sphere $x² + y² + z² = r$ that would have a tangent plane with $x = \sqrt{r}$ if $y=0$ and $z=0$ and from that reasoning I derive that the radius can't simply be $r$ for a plane. Even though I understand that I am missing something in my reasoning.
The plane is really a tangent to the ellipsoid. The point you're missing is $(1,1,1)$, which is the tangential point between the surface and the plane. See $1+2+3=6=1^2+2(1)^2+3(1)^2$. Check also $(6,0,0)$ and $(0,3,0)$ lie on the plane.
If you set $y=0,z=0$ in the equation of the plane, you'll end up with some other points on the plane. If you plug a point into the equation of the ellipsoid, it gives you the information whether the point is inside the ellipsoid or not. From you example, you set $y=0,z=0$, the ellipsoid gives you $x=\sqrt{6}$ because there's a point $(\sqrt{6},0,0)$ on the surface, but not on the plane. $(6,0,0)$ is on the plane but not the surface. This has no implication whether the plane is not a tangent plane and so on. The only moment you have two equations equal to each other, is at the point $(1,1,1)$.