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If $$2^{x}=3^{y}=6^{-z}$$, then $$\left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)$$ is equal to
$$2^{x}=3^{y}=6^{-z}$$ = K
X = $$\log_2k$$
Y = $$\log_3k$$
Z = -($$\log_6k$$)
Hence, $$\left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)$$ = $$\log_k2$$ + $$\log_k3$$ - $$\log_k6$$
= $$\log_k6$$ - $$\log_k6$$
= 0.