Closest point on a plane to a surface. And vice versa. Find the point on $z=1-2x^2-y^2$ closest to $2x+3y+z=12$ using Lagrange multipliers.
Point on surface closest to a plane using Lagrange multipliers
Although the methods used in the answers are helpful and do work, my professor told me that the way he wants us to do this problem is to recognize that the distance will be minimized when the normal vector to one point on the surface is parallel to the normal vector of the plane. To find the normal vector to the surface at a point $(x,y,z)$ we write our function implicitly and take the gradient of the new function (gradient is parallel to level curve):
$$G(x,y,z,)=z+2x^2+y^2=1$$
$$\nabla G=\langle4x,2y,1\rangle$$
The normal vector of the plane is,
$$\langle 2,3,1 \rangle$$
Hence we have,
$$\langle 4x,2y,1 \rangle=\lambda \langle 2,3,1 \rangle$$
And that,
$$z+2x^2+y^2=1$$
Question $1$:

The part I don't understand is the claim that the distance will be minimized when the normals are parallel. And how is this using Lagrange multipliers, May someone please explain. 

Question $2$
Find the point on $2x+3y+z=12$ closest to $z=1-2x^2-y^2$ using Lagrange multipliers.
I suppose we have use the fact that the normal of the plane is $\langle 2,3,1 \rangle$ with the fact that the point closest to the plane on the surface is $(\frac{1}{2},\frac{3}{2},-\frac{7}{4})$  to come up with the equation of the line that goes through our two closest points in terms of $t$ and then substitute values of $x$, $y$, $z$ into our plane equation to come up with the point on the plane. But again, I don't see where Lagrange multipliers comes into play.
 A: Call $\zeta$ the coordinates of the point on the surface $g(\zeta) = 0$, and $\xi$ the coordinates on the plane $h(\xi) = 0$. The problem you are solving is then 
\begin{eqnarray}
\mbox{minize: }\quad f(\zeta, \xi) &=& \frac{1}{2}\sum_\alpha(\zeta_\alpha - \xi_\alpha)^2 \\
\mbox{suject to:}\quad g(\zeta) &=& 0 \\
h(\xi) &=& 0,
\end{eqnarray}
which can be solved by minimizing the function
$$
F(\zeta,\xi, \lambda, \mu) = f(\zeta,\xi) + \lambda g(\zeta) + \mu h(\xi)
$$
with $\lambda$ and $\mu$ being the Lagrange multipliers. This is just done by calculating the gradient of $F$ w.r.t. all of its arguments ($\zeta$, $\xi$, $\lambda$ and $\mu$) and making it equal to zero,
\begin{eqnarray*}
\nabla_\zeta f (\zeta, \xi)+\lambda \nabla g(\zeta) &=& 0 \\
\nabla_\xi f(\zeta,\xi)+\mu \nabla h(\xi) &=& 0 \\
g(\zeta) &=& 0 \\
h(\xi) &=& 0
\end{eqnarray*}
The important thing to realize here is that 
$$
\frac{\partial f}{\partial\zeta_\alpha} = \frac{\partial f}{\partial\xi_\alpha} = (\zeta_\alpha - \xi_\alpha)
$$
And then, by substracting the first two equations, we arrive to the conclusion:
$$
\lambda \nabla g = - \mu \nabla h
$$
or equivalently: the normals are parallel. And I believe that answers your question. 
BTW, at the beginning of your post you said 'parallel' and later in the box 'perpendicular'. Please fix that
A: Answering to "I don' understand..claim that the distance will be minimized when the normals are parallel".
On a pure intuitive basis (not pretending to be rigorous) consider a point on the given surface, and the tangent plane to the surface at that point.
In a small domain around the point, the surface is "approximated" by the tangent plane $t$.
If this is not parallel to the other given plane $p$, then there will be a point in that small domain which is closer to $p$ than the original.
Move in that point, repeat the conceptual operation, and you will end (if exists) onto a point on the surface, where the tangent plane is parallel to $p$. 
