The equation presented in the image is:

$2^x = 5^y = (10)^{-z}$ and the task is to find:

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

Step-by-step Solution:

1. Start by expressing $10^{-z}$ in terms of its prime factorization: $10 = 2 \times 5 \quad \text{so} \quad 10^{-z} = 2^{-z} \times 5^{-z}.$ This gives us: $2^x = 5^y = 2^{-z} \times 5^{-z}.$

2. Set the first equation $2^x = 2^{-z}$, which implies: $x = -z.$

3. Similarly, from $5^y = 5^{-z}$, we get: $y = -z.$

4. Now substitute these results into the expression we need to evaluate: $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}.$ Since $x = -z$ and $y = -z$, we have: $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{-z} + \frac{1}{-z} + \frac{1}{z}.$

5. Simplifying: $\frac{1}{-z} + \frac{1}{-z} + \frac{1}{z} = -\frac{1}{z} - \frac{1}{z} + \frac{1}{z} = -\frac{1}{z}.$

So, the final answer is:

$\boxed{0}.$

Would you like further clarification on any step or more details? Here are some related questions:

1. How do you derive $10^{-z}$ as $2^{-z} \times 5^{-z}$?
2. What is the significance of expressing numbers as powers in such problems?
3. Can you explain why $x = -z$ and $y = -z$ hold true for this problem?
4. How would the answer change if the bases of the exponentials were different?
5. Can this problem be solved using logarithms?

Tip: Understanding how to manipulate exponents is critical in solving equations involving multiple bases like in this question.