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 $A+B+C=100$. This has $\binom{99}{2}=4851$ solutions.