An Exponent problem

$2^{x} = 3^{y} = 6^{-z}$

$2 = 6^{\frac{-z}{x}}$

$3 = 6^{\frac{-z}{y}}$

$2 × 3 = 6^{\frac{-z}{x}} × 6^{\frac{-z}{y}}$

$6^{1} = 6^{\frac{-z}{x}} × 6^{\frac{-z}{y}}$

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

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

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

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

let 2^x=3^y=6^-z=km,2=k^1/x,3=k^1/y,6=k^1/-z now, 2 3=6,k^1/x k^1/y=k^1/-z,k^(1/x+1/y)=k^1/z therefore,1/x+1/y=1/-z,1/x+1/y+1/z=0