I don't have time to work out all the details in an answer, but here's a quick starter.
The key idea behind Lagrange multipliers is that when two surfaces are tangent to each other, their normal vectors at that point are parallel. In this case, you want to find when the surface $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$ is closest to the origin, that is, when it is tangent to a sphere of some radius.
Note first that if $F(x,y,z) = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$, the surface is equal to $F^{-1}(1)$, so the normal field is given by the gradient $\nabla F$.
It's easier to use the energy function $d^2(x,y,z) = x^2 + y^2 + z^2$, because square roots are silly. To find where distance from the origin is minimized, find the gradient of the distance function, $\nabla (d^2)$, and find where it points the same direction as $\nabla F$:
$$\nabla(d^2) = \lambda \nabla F,$$
subject to $F = 1$.
The squared distance from $(x,y,z)$ to $(2,0,-1)$ is $(x-2)^2 + (y-0)^2 + (z+1)^2$.
(By the way, distance to a plane can also be calculated just using Cauchy-Schwarz; but I guess you've had the method specified for you, so it doesn't really matter.)
Best Answer
A simpler approach for computing the distance between these objects: the given plane is orthogonal to the vector $(2,3,1)^T$, so the point(s) on the surface of minimal distance from the plane are the ones for which the tangent plane of the surface at such point(s) is orthogonal su $(2,3,1)^T$. If for some $k$ $$\left\{\begin{array}{rcl} 2x^2+y^2+z &=& 1 \\ 2x+3y+z &=& k \end{array}\right. $$ has exactly one solution, such a solution is a point of minimal distance. By eliminating $z$, we are looking for the values of $k$ such that $$ 2x^2-2x+y^2-3y = 1-k $$ has exactly one solution. By completing the squares, it is trivial that the only point of minimal distance occurs at $(x,y)={\left(\frac{1}{2},\frac{3}{2}\right)}$, from which $(x,y,z)=\color{red}{\left(\frac{1}{2},\frac{3}{2},-\frac{7}{4}\right)}$ and $$ d(\text{plane},\text{surface}) = \frac{\left|2\cdot\frac{1}{2}+3\cdot\frac{3}{2}-\frac{7}{4}-12\right|}{\sqrt{3^2+2^2+1^2}}=\color{red}{\frac{33}{4\sqrt{14}}}.$$