The crucial observation is: if $a$ is too big, then $\frac{1}{a}$ is too small (and analogously for $b$ and $c$). Hence, the potential solutions involve natural numbers that have upper bounds.
It's not the most elegant solution, but since there are only three variables, you can do a case bash. WLOG, let's assume that $0 < a \leq b \leq c$, and we obtain all solutions by permutations of any solution $(a, b, c)$.
Notice that $a > 1$. So, consider $a = 2$. Then
$$
\frac{1}{b} + \frac{1}{c} = \frac{3}{10}.
$$
Since $\frac{1}{b} \leq \frac{3}{10}$, we must have $b \geq \frac{10}{3} > 3$. Now, since we've assumed that $b \leq c$, we have $\frac{1}{b} \geq \frac{1}{c}$, so
$$
\frac{2}{b} \geq \frac{3}{10},
$$
hence $3b \leq 20$ or $b \leq \lfloor \frac{20}{3} \rfloor = 6$.
Putting these bounds together, $b \in \{4, 5, 6\}$.
With $b = 4$, we have $c = 20$, so $(a, b, c) = (2, 4, 20)$ is a solution. With $b = 5$, we have $c = 10$, so $(a, b, c) = (2, 5, 10)$ is another solution. With $b = 6$, $c \notin \mathbb{N}$, so we've found all solutions for $a=2$.
Now, consider $a = 3$. Then,
$$
\frac{1}{b} + \frac{1}{c} = \frac{7}{15}.
$$
Similar arguments yield $7b \leq 30$, or $b \in \{3, 4\}$. Unfortunately, neither of these possibilities yields a solution with $c \in \mathbb{N}$.
Finally, consider $a \geq 4$, so
$$
\frac{1}{b} + \frac{1}{c} \leq \frac{11}{20}.
$$
We have $b \geq 2$ and $11b \leq 40$ or $b \leq \frac{40}{11} < 4$, but then none of $a$, $b$, or $c$ is a multiple of $5$, so we can't have any more solutions.
Hence, there are $3! \cdot 2 = 12$ solutions, consisting of the $3! = 6$ permutations of each of $(2, 4, 20)$ and $(2, 5, 10)$.
