Distance between hyperplane and ellipsoid How would you use Lagrange multipliers to find the following?

Minimise the distance between ellipsoid $$x^2 + y^2 + \frac{10}{9} z^2 - 10^{10}=0$$ and plane $$x + y + z - 10^8 = 0$$

I computed the normal vector to the plane which is $(1,1,1)$ but I really don't know how to proceed.
 A: I elaborate the ideas that has been expressed in other two answers to show that they are, actually, the same.
A distance from a point $P\colon (x,y,z)$ to a plane $ax+by+cz+d=0$ is given by
$$
d(x,y,z)=\frac{|ax+by+cz+d|}{\sqrt{a^2+b^2+c^2}},
$$
thus, our problem, as the first answer suggests, is to minimize $d(x,y,z)$ subject to $P\colon(x,y,z)$ belonging to the ellipsoid $x^2+y^2+kz^2=m$. Before applying the Lagrange multipliers method we would like to simplify the problem first to make the resulting necessary condition look easier. Observe that to minimize $d$ is equivalent to minimizing $|ax+by+cz+d|$, which, in turn, is equivalent to minimizing the smooth function
$$
f(x,y,z)=(\underbrace{ax+by+cz+d}_{=h})^2=h^2(x,y,z).
$$
Hence, the simplified problem is
$$
\min f(x,y,z)\quad\text{subject to }g(x,y,z)=x^2+y^2+kz^2-m=0.
$$
Write down the necessary condition for the optimum
$$
\nabla f\,\|\,\nabla g\quad\Leftrightarrow\quad 
2h(x,y,z)\begin{bmatrix}a\\b\\c\end{bmatrix}\ \|\ 2\begin{bmatrix}x\\y\\kz\end{bmatrix}\quad\Leftrightarrow\quad h(x,y,z)=0\ \text{ or }\  \begin{bmatrix}x\\y\\kz\end{bmatrix}=\lambda\begin{bmatrix}a\\b\\c\end{bmatrix}.
$$
The first equality takes care of the case where the plane intersects the ellipsoid and the second one means that the tangent plane at a possible optimum is parallel to the given plane. This is the geometric approach in the second answer.
To finish the problem you need


*

*to check if there is an intersection between the plane and the ellipsoid. Then the minimal distance is zero.

*to substitute, if there is no intersection, the candidate point 
$$
\begin{bmatrix}x\\y\\z\end{bmatrix}=\lambda\begin{bmatrix}a\\b\\c/k\end{bmatrix}
$$
to the equation $x^2+y^2+kz^2=m$, find (two) $\lambda$ and pick the candidate with the smallest $f(x,y,z)$.

A: The function that you need to minimize is the distance to the plane, call it $d(x, y, z)$. The constraint is that $(x,y,z)$ lies on the ellipsoid, so if $g(x,y,z)=0$ is the equation of the ellipsoid, the following equalities will hold for some $\lambda$ and the minimum:
$$
\nabla (d-\lambda g)(x,y,z)=0
$$
$$
g(x,y,z)=0
$$
A: Here is a geometric solution:
The point on the ellipsoid where minimum distance is attained is where the normal vector of ellipsoid at that point is parallel to normal vector of the plane.
the outer normal vector at point $(x,y,z)$ of ellipsoid is $(x,y, \frac{10}{9}z)$
this must be a multiple of $(1,1,1)$ which means $(x,y, \frac{10}{9}z) = (t,t,t )$ and satisfies the ellipsoid. now by plug in back to the equation of ellipsoid you get $t= \frac{2 \times10^5 \times \sqrt 7}{3}$.
Therefore you can get both point and minimum distance. 
