In the following alpha-numeric puzzle, each letter represents a different non-zero digit. What are all possible values for $b+e+h$?

 $ \begin{tabular}{cccc} &a&b&c \\ &d&e&f \\ + & g&h&i \\ \hline 1&6&6&5 \end{tabular}$

[i]Proposed by Eugene Chen[/i]
1. We are given the alpha-numeric puzzle where each letter represents a different non-zero digit:
   \[
   \begin{array}{cccc}
   & a & b & c \\
   & d & e & f \\
   + & g & h & i \\
   \hline
   1 & 6 & 6 & 5 \\
   \end{array}
   \]
   We need to find all possible values for \( b + e + h \).

2. First, we note that the sum of the three numbers must equal 1665. This means:
   \[
   100a + 10b + c + 100d + 10e + f + 100g + 10h + i = 1665
   \]

3. We observe that the last digit of the sum (units place) is 5. Therefore, the sum of the units digits \( c + f + i \) must end in 5. Since \( c, f, i \) are distinct non-zero digits, the only possible sum that ends in 5 is 15. This is because:
   \[
   \text{Maximum sum} = 9 + 8 + 7 = 24 \quad (\text{which is less than 25})
   \]
   \[
   \text{Minimum sum} = 1 + 2 + 3 = 6 \quad (\text{which is greater than 5})
   \]
   Therefore, \( c + f + i = 15 \).

4. Similarly, the sum of the tens digits \( b + e + h \) must also end in 6 (since the tens place of 1665 is 6). However, since we are dealing with distinct non-zero digits, we need to consider the carry from the units place. Given that \( c + f + i = 15 \), there is a carry of 1 to the tens place.

5. Therefore, the sum of the tens digits \( b + e + h \) plus the carry from the units place must end in 6. This means:
   \[
   b + e + h + 1 = 16 \implies b + e + h = 15
   \]

6. To verify, we can consider an example where the digits are distinct and non-zero. For instance, if we take:
   \[
   123 + 564 + 978 = 1665
   \]
   Here, \( b = 2 \), \( e = 6 \), and \( h = 7 \). Thus:
   \[
   b + e + h = 2 + 6 + 7 = 15
   \]

Conclusion:
\[
\boxed{15}
\]