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 use of two Lagrangian multipliers to solve the problem seems to be entirely sound, and so I have really nothing to add there. But it's worth noting that you can actually get away with just one such multiplier!
Since $x^2+y^2=z^2$, we have $f(x,y,z)=x^2+y^2+z^2=2x^2+2y^2$. Then both this objective function and the constraint $g(x,y)=x+2y=6$ depend only $(x,y)$; all that remains is $x^2+y^2=z^2$, which doesn't constrain $x,y$ unless $z$ is known. So we can set aside this equation, and what we have is an optimization problem in $(x,y)$. Introducing a Lagrangian multiplier $\lambda$, we demand that $\nabla f = (4x,4y)=\lambda \nabla g = (\lambda,2\lambda)$. This implies $\lambda = 2y=4x$ and so $(x,y)=(\frac65,\frac{12}5).$ Then $z=\pm \sqrt{x^2+y^2}=\pm\frac{6}{\sqrt{5}}$ and the minimum distance is $\frac{12}{\sqrt{5}}.$
Best Answer
The variables are $x$, $y$ and $z$ .
The objective function is minimizing the distance to the origin $f(x,y,z)=x^2 +y^2+z^2$.
The constraints are $g(x,y,z)=-1-z \leq 0$ and $h(x,y,z)=x^2+y^2+z-1=0$.
If you use KKT conditions, it will be easy to solve the problem.