Let's review this in $\mathbb R^3$:
Consider any two points $(x,y,z)$ and $(a,b,c)$ in $\mathbb R^3$. The distance between them is
$D=\sqrt { (x-a)^{2}+(y-b)^{2}+(z-c)^{2}}$.
Now suppose $(a,b,c)$ is fixed and that $(x,y,z)$ lie in the plane $Ax+By+Cz=E$.
Your problem is to find the point(s) on the plane such that $D$ is minimum. One way to do it is by observing that at least one of the coefficients $A,B$ or $C$ is non-zero, so that you can solve for the corresponding variable, substitute into $D$ and use the second derivative test for functions of two variables.
If you are going to use Lagrange, then you use the fact that the equation of the plane provides the constraint equation; that is,
$g(x,y,z)=Ax+By+Cz-E$ so that, according to the method, you construct
$D(x,y,z,\lambda)=\sqrt { (x-a)^{2}+(y-b)^{2}+(z-c)^{2}}-\lambda g(x,y,z)$ which is just
$D(x,y,z,\lambda)=\sqrt { (x-a)^{2}+(y-b)^{2}+(z-c)^{2}}-\lambda (Ax+By+Cz-E)$.
and then you solve the system
$D_{x}(x,y,z,\lambda)=0;D_{y}(x,y,z,\lambda)=0;D_{z}(x,y,z,\lambda)=0;D_{\lambda}(x,y,z,\lambda)=0$
which gives you the possible extreme values for $D$.
For the corresponding problem in $\mathbb R^n$, simply note that here
$D(x_{1},\cdots ,x_{n})=\sqrt {\sum_{i=1}^{n}(x_{i}-a_{i})^{2}}$ and
$g(x_{1},\cdots ,x_{n})=\sum_{i=1}^{n}A_{i}x_{i}-E$
so you construct $D(x_{1},\cdots ,x_{n},\lambda )$ as before but now you have $n+1$ equations in $n+1$ unknowns to solve.
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
You have 5 equations and 5 unknowns and you have to find all solutions to the system. Some solutions will correspond to local maxima, other to local minima, and the remaining ones to local inflection points. Then, among the solutions that you found, you have to evaluate f(x) and see which one has highest value (or use a Hessian test which is not necessary here).
In the very first step of solving the system, you assumed that $y\ne0$ and so you ended up dropping that solution. If you considered the $y=0$ case separately, you would have recovered that solution.