If $2^{x} = 3^{y} = 6^{-z}$, then $(\frac{1}{x} + \frac{1}{y} + \frac{1}{z})$ is equal to

Let's call the common value $a$.

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

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Converting to logarithmic form:

From $2^x = a$, we get $x = \log_2(a)$

From $3^y = a$, we get $y = \log_3(a)$

From $6^{-z} = a$, we get $z = -\log_6(a)$

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Finding the reciprocals:

$\dfrac{1}{x} = \dfrac{1}{\log_2(a)}$

$\dfrac{1}{y} = \dfrac{1}{\log_3(a)}$

$\dfrac{1}{z} = \dfrac{1}{-\log_6(a)} = -\dfrac{1}{\log_6(a)}$

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Using the change of base formula to standardize:

$\log_3(a) = \dfrac{\log_2(a)}{\log_2(3)}$

$\log_6(a) = \dfrac{\log_2(a)}{\log_2(6)}$

This gives:

$\dfrac{1}{y} = \dfrac{\log_2(3)}{\log_2(a)}$

$\dfrac{1}{z} = -\dfrac{\log_2(6)}{\log_2(a)}$

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Adding the fractions:

$\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{\log_2(a)} + \dfrac{\log_2(3)}{\log_2(a)} - \dfrac{\log_2(6)}{\log_2(a)}= \dfrac{1 + \log_2(3) - \log_2(6)}{\log_2(a)}$

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Since $\log_2(6) = \log_2(2 \times 3) = \log_2(2) + \log_2(3) = 1 + \log_2(3)$:

$\dfrac{1 + \log_2(3) - \log_2(6)}{\log_2(a)} = \dfrac{1 + \log_2(3) - (1 + \log_2(3))}{\log_2(a)} = \dfrac{0}{\log_2(a)} = 0$

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$\therefore \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0$