3-variable 3-equations problem 
For positive reals $x,y,z$, if $xy+yz+xz=\sqrt{\frac pq}$, given
$$x^2+xy+y^2=2$$
$$y^2+yz+z^2=1$$
$$z^2+zx+x^2=3$$
find $p-q$, where $p$ and $q$ are relatively prime integers.

I tried grouping the required sum by adding all the equations but it led me to nowhere. I tried applying basic AM-GM inequality to conclude that the sum is less than or equal to $2$, but this is of no use since $x,y,z$ are positive reals, not just positive integers.
I have exhausted all my ideas. Any hint/help is appreciated!
 A: Actually there is a geometric solution to this. Consider a point $P$ in $\mathbb{R}^2$ and points $A$, $B$, $C$ with $PA=x$, $PB=y$, $PC=z$ and $\angle APB=\angle APC=\angle BPC=120^{\circ}$.
If you apply the cosine theorem you get that the triangle $\triangle ABC$ has sides $BC=1$, $AC=\sqrt{3}$, $AB=\sqrt{2}$. Notice that it is a right triangle since $AC^2=BC^2+AB^2$.

Computing the area in two ways $xy+yz+zx=\cfrac{4}{\sqrt{3}}\cdot \operatorname{Area}(\triangle ABC)=\cfrac{2\sqrt{2}}{\sqrt{3}}$
As a remark the numbers are quite incidental here; you can pick your favorite numbers for the system and do the same construction and instead use Herron formula for the area to compute $xy+yz+zx$.
A: If you like a more algebraic answer, consider the following linear combination of the three equations in the order written above $(x-y)(1)+(y-z)(2)+(z-x)(3)$. The LHS of the resultant equation vanishes and we get a linear relation:
$$ 2(x-y)+(y-z)+3(z-x)=0\iff z=\frac{x+y}{2}$$
Under this light we see that $\Pi=xy+xz+yz=6z^2-2$ because in fact equation $(1)$ can be rewritten in the form
$$x^2+xy+y^2-2=0\iff xy=4z^2-2$$
Thus we only need to compute $z^2$. In the following denote $\rho=\frac{x-y}{2}$. Then consider the combination $(3)-(2)$ we get
$$x^2-y^2+(x-y)z=2\iff z\rho=\frac{1}{3}$$
and finally we note that we can write
$$x^2+xy+y^2=(x+y)^2-xy=3z^2+\rho^2=2$$
The last two equations can be solved by substitution of the 1st into the 2nd, which yields the equation
$$3z^4-2z^2+\frac{1}{9}=0\iff 6z^2=2\pm\sqrt{\frac{8}{3}}$$
which since $\Pi>0$ yields immediately
$$\Pi=\sqrt{\frac{8}{3}}\Rightarrow p=8~,~q=3$$
and thus $p-q=5$.
