Skorzystamy z założenia:

$\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=0$  

$\frac{yz}{xyz}+\frac{xz}{xyz}-\frac{xy}{xyz}=0$  

$\frac{yz+xz-xy}{xyz}=0$  

$yz+xz-xy=0$  

$xy=yz+xz$   

oraz

$x+y-z=13\mid {}^2$

${\left(x+y-z\right)}^2=169$  

${\left(\left(x+y\right)-z\right)}^2=169$  

${\left(x+y\right)}^2-2\cdot \left(x+y\right)\cdot z+z^2=169$  

$x^2+2xy+y^2-2xz-2yz+z^2=169$  

$x^2+2xy-2xz-2yz+y^2+z^2=169$  

$x^2+2\left(xy-xz-yz\right)+y^2+z^2=169$  

$x^2+2\cdot \left(yz+xz-xz-yz\right)+y^2+z^2=169$  

$x^2+2\cdot 0+y^2+z^2=169$  

$x^2+y^2+z^2=169$  

Kodujemy:

$169$