How to solve the equations $\sqrt{x-3}+\sqrt{y-3}=\sqrt{y-12}+\sqrt{z-12}=\sqrt{z-27}+\sqrt{x-27}=12$ 
Let $x,y,z\in R$, 
  and
  $$\begin{cases}
\sqrt{x-3}+\sqrt{y-3}=12\\
\sqrt{y-12}+\sqrt{z-12}=12\\
\sqrt{z-27}+\sqrt{x-27}=12
\end{cases}$$
  Find the $x,y,z$.

My try: I want use The geometry to solve it. ( Norbert have solved it)
and I think this problem have algebra methods.Thank you 

we only find this $x,y,z$,
then we have
$$OP^2=ON^2+NP^2-ON\cdot NP=39$$
and 
$$\dfrac{OP}{\sin{A}}=x\Longrightarrow \sqrt{x}=2\sqrt{13}$$
and use the same methods we easy to find $y,z$
I think of seeing algebra methods? maybe use if $x>y$,then use
$$\sqrt{x-3}+\sqrt{y-3}=\sqrt{y-12}+\sqrt{z-12}<\sqrt{x-12}+\sqrt{z-12}?$$
then I can't, 
Thank you 
 A: Actually your geometric idea is wonderful! But I had to correct your shape:
$NI || BC,NJ||BC$, so $\angle PIJ = \angle NJA=60$, So $NI=6,NJ=2,IJ=8,NH=2$ therefore $x=(4\sqrt 3)^2+2^2=52$. I am sure you can continue!$y=28,z=76$
EDIT: This solution is for a time that OP said he/she tried the geometric idea but without any success. Then OP edited the post and said he/she succeeded through geometric method.
A: Clearly $OAP=60^\circ$ $ONP=120^\circ$, so from triangle $ONP$ via cosine theorem you can find $OP^2=39$. Since $AONP$ have two right angles it is inscribed in some circle. Clearly $AN$ its diameter. Now apply sine theorem for trianle $ONP$ which is inscribed in the same circle.
A: Here's another algebraic solution, similar to Norbert's but taking a somewhat different tack.
For all the square roots to be real, we must have $x,z\ge27$ and $y\ge12$, so let's let
$$\begin{align}
x&=27+u^2\\
y&=12+v^2\\
z&=27+w^2\\
\end{align}$$
with $u,v,w\ge0$. Then the equations become
$$\begin{align}
\sqrt{u^2+24}+\sqrt{v^2+9}&=12\\
v+\sqrt{w^2+15}&=12\\
u+w&=12\\
\end{align}$$
Thus
$$w=12-u \quad\text{and}\quad v=12-\sqrt{(12-u)^2+15}$$
so the equation to be solved is
$$\sqrt{u^2+24}+\sqrt{\left(12-\sqrt{(12-u)^2+15}\right)^2+9}=12$$
This "simplifies" first to
$$\left(12-\sqrt{u^2-24u+159}\right)^2+9=\left(12-\sqrt{u^2+24}\right)^2$$
which boils down to
$$\sqrt{u^2-24u+159}=\sqrt{u^2+24}-(u-6)$$
This squares and simplifies to
$$2(u-6)\sqrt{u^2+24}=u^2+12u-99$$
and this squares to produce a quartic that factors as
$$3(u-5)(u^3-19u^2+3u+423)=0$$
We thus have the solution $u=5$, which leads to $x=52$, $y=28$ and $z=76$, as Norbert found.  The cubic factor looks unpleasant, but we don't have to worry about any of its roots that are less than $0$ or greater than $12$, since those do not satisfy the non-negativity conditions on $u$ and $w$.  Now if we go back to the original equation and let
$$f(u)=\sqrt{u^2+24}+\sqrt{\left(12-\sqrt{(12-u)^2+15}\right)^2+9}-12$$
we have
$$f'(u)={u\over\sqrt{u^2+24}}+{12-\sqrt{(12-u)^2+15}\over\sqrt{\left(12-\sqrt{(12-u)^2+15}\right)^2+9}}\cdot{(12-u)\over\sqrt{(12-u)^2+15}}$$
which is clearly positive for $1\le u\le 12$, so we don't have to worry about any roots of the cubic in that range either.  Finally, it's easy to see that the cubic decreases from $423$ to $408$ in the interval $0\lt u\lt1$.  Thus $u=5$ is the only solution.
