Solution:
Let $x = a + b + c$ and $y = d + e + f$. Then the equation becomes
$$x + 2y = 15$$
where $a, b, c, d, e, f$ are positive integers, so $x \geq 3$ and $y \geq 3$.

Let us solve for all possible integer values of $y$ such that $y \geq 3$ and $x = 15 - 2y \geq 3$.

We have:
$$15 - 2y \geq 3 \implies 2y \leq 12 \implies y \leq 6$$
So $y$ can be $3, 4, 5, 6$.

For each $y$, $x = 15 - 2y$.

For each $y$, the number of positive integer solutions to $d + e + f = y$ is $\binom{y-1}{2}$.
For each $x$, the number of positive integer solutions to $a + b + c = x$ is $\binom{x-1}{2}$.

So the total number is:
$$\sum_{y=3}^6 \binom{y-1}{2} \binom{15-2y-1}{2}$$
Let us compute each term:

For $y = 3$:
$\binom{2}{2} = 1$, $x = 9$, $\binom{8}{2} = 28$
So $1 \times 28 = 28$

For $y = 4$:
$\binom{3}{2} = 3$, $x = 7$, $\binom{6}{2} = 15$
So $3 \times 15 = 45$

For $y = 5$:
$\binom{4}{2} = 6$, $x = 5$, $\binom{4}{2} = 6$
So $6 \times 6 = 36$

For $y = 6$:
$\binom{5}{2} = 10$, $x = 3$, $\binom{2}{2} = 1$
So $10 \times 1 = 10$

Add them up:
$$28 + 45 + 36 + 10 = 119$$

Answer: $119$