Suppose f is a function satisfying $$f(x + y) = f(x) + f(y)$$ for all $$x, y \in \mathbb{N}$$ and $$f(1) = \frac{1}{5}$$. If $$\sum_{n=1}^{m} \frac{f(n)}{n(n+1)(n+2)} = \frac{1}{12}$$ then m is equal to ______.

Given the function satisfies the property:

$$f(x + y) = f(x) + f(y)$$

$$f2) = f(1) + f(1) = 2f(1)$$

$$f(3) = f(2) + f(1) = 3f(1)$$

$$ \implies f(n) = nf(1) $$

We are given that $$f(1) = \frac{1}{5}$$, so we can substitute this to find $$f(n)$$:

$$f(n) = \frac{n}{5}$$

Now, substitute this expression for $$f(n)$$ into the given summation:

$$\sum_{n=1}^{m} \frac{f(n)}{n(n+1)(n+2)} = \sum_{n=1}^{m} \frac{\frac{n}{5}}{n(n+1)(n+2)}$$

$$S_m= \frac{1}{5} \sum_{n=1}^{m} \frac{1}{(n+1)(n+2)}$$

The general term in summation is given by 

$$T_n = \frac{1}{(n+1)(n+2)} = \frac{1}{n+1} - \frac{1}{n+2}$$

Then $$S_m$$ simplifies to

$$S_m = \frac{1}{5} (\frac{1}{2} - \frac{1}{m+2}) $$

$$S_m = \frac{m}{10(m+2)} = \frac{1}{12}$$

$$6m= 5m + 10$$

$$ m = 10 $$