Given the equation $2^x = 3^y = 6^z = k$, where $k$ is some constant, we need to find the value of $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$.

1. Express in terms of $k$:

   • $2^x = k$ implies $x = \frac{\log k}{\log 2}$
   • $3^y = k$ implies $y = \frac{\log k}{\log 3}$
   • $6^z = k$ implies $z = \frac{\log k}{\log 6}$

2. Find $\frac{1}{x}$, $\frac{1}{y}$, and $\frac{1}{z}$:

   • $\frac{1}{x} = \frac{\log 2}{\log k}$
   • $\frac{1}{y} = \frac{\log 3}{\log k}$
   • $\frac{1}{z} = \frac{\log 6}{\log k}$

3. Sum up $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$:

   $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{\log 2}{\log k} + \frac{\log 3}{\log k} + \frac{\log 6}{\log k}$

   This can be factored as:

   $\frac{\log 2 + \log 3 + \log 6}{\log k}$

   Using the logarithm property $\log ab = \log a + \log b$ and noting that $\log 6 = \log (2 \times 3) = \log 2 + \log 3$:

   $\log 2 + \log 3 + \log 6 = \log 2 + \log 3 + (\log 2 + \log 3) = 2(\log 2 + \log 3)$

4. Conclusion:

   Since the numerator simplifies to $2(\log 2 + \log 3)$, and the denominator is $\log k$:

   $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{2(\log 2 + \log 3)}{\log k}$

   Given that $\log k = \log 6^z = z\log 6 = z(\log 2 + \log 3)$, the expression simplifies further:

   $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{2(\log 2 + \log 3)}{z(\log 2 + \log 3)} = \frac{2}{z}$

Finally, substitute ( z