Calculate the sum $x+y+z$ If positive real numbers $x,y,z$ satisfying the conditions
$$x^2+xy+y^2=25, y^2+yz+z^2=49, z^2+zx+x^2=64$$ 
then calculate the sum $x+y+z.$
My attempt:
By subtracting equalities are obtained $$(z-x)(x+y+z)=24, (x-y)(x+y+z)=15.$$
And here I stopped.    )-:


$\angle ADB=\angle CDB= \angle ADC=120^{\circ}$
$D -$ Torricelli Point
$$S_{ADB}+S_{CDB}+S_{ADC}=S_{ABC}$$
$$\frac 12 xy \sin 120^{\circ}+\frac 12 zy \sin 120^{\circ}+\frac 12 xz \sin 120^{\circ}=\sqrt{10 \cdot 3 \cdot 3 \cdot 2}$$
$$\frac {\sqrt 3}{4} xy +\frac {\sqrt 3}{4} zy +\frac {\sqrt 3}{4} xz=6\sqrt{5}$$
$$\frac {\sqrt 3}{4}(xy+yz+zx)=6\sqrt{5}$$
$$xy+yz+zx=\frac{24\sqrt 5}{\sqrt3}$$
$$x^2+xy+y^2+y^2+yz+z^2+z^2+zx+x^2=25+64+49$$ 
$$2(x^2+y^2+z^2)+4xy+4yz+4zx=138+3 \cdot \frac{24\sqrt 5}{\sqrt3}$$
$$(x+y+z)^2=69+12\sqrt{15}$$
$$$$
