, 2017

Solution:
Note that $6a$ must be a multiple of $5$, so $a$ must be a multiple of $5$. Similarly, $b$ must be a multiple of $3$, and $c$ must be a multiple of $2$.

Set $a = 5A$, $b = 3B$, $c = 2C$. Then the equation reduces to
$$6a + 10b + 15c = 6(5A) + 10(3B) + 15(2C) = 30A + 30B + 30C = 30(A + B + C) = 3000$$
So $A + B + C = 100$.

The number of ordered triples of positive integers $(A, B, C)$ such that $A + B + C = 100$ is
$$\binom{99}{2} = 4851.$$