Shortest distance between ellipsoid and plane Find the shortest distance between the points $A = (x_1,y_1,z_1)$ and $B = (x_2,y_2,z_2)$ if $A$ lies on the plane $x+y+z=2a$ and $B$ lies on the ellipsoid $$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$$

The prameterization of an ellipsoid is 
$$x=a\cos\theta\sin\phi \\ 
  y=b\cos\theta\cos \phi \\ 
  z=c\sin\theta$$
distance of a point from a $x+y+z-2a=0$ is given by $$l=\frac{x+y+z-2a}{\sqrt3} \\ \implies l=\frac{a\cos\theta\sin\phi+b\cos\theta\cos\phi+c\sin\theta-2a}{\sqrt 3}$$
For minimum distance $$\frac{\partial l}{\partial\theta}=0;\quad\frac{\partial l}{\partial\phi}=0$$
