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It's easy to see that $1 \leq a \leq 12$ since for $a = 13$ the left side of the equation $29a + 30b + 31c = 366$ is greater than $366$ . Using that reasoning we get that $1 \leq b \leq 12$ and $1 \leq c \leq 11$ . Then:

$a \in [1, 12] \subset \mathbb{Z} \Rightarrow a = \big\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\big\} \\ b \in [1, 12] \subset \mathbb{Z} \Rightarrow b = \big\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\big\} \\ c \in [1, 12] \subset \mathbb{Z} \Rightarrow c = \big\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\big\}$

$29a + 30b + 31c = 366 \\ \Rightarrow 30a - a + 30b + 30c + c = 366 \\ \Rightarrow 30(a+b+c) = 366 +a -c \\ \Rightarrow a+b+c = 12 + \frac{6+a-c}{30} \\ \stackrel{\text{let } a-c = 30k - 6}{\Longrightarrow} a+b+c = 12 +k$

Now we observe that $k = 0$ since for either $k \geq 1$ or $k \leq -1$ there are none of the given values of $a, c$ that satisfy the equation $a - c = 30k -6$

Ummmm if you notice, 366 is the no. of days in a leap year, and 29, 30 and 31 are the corresponding no. of days in each month. Hence a+b+c is equal to the total no. of months i.e. 12!!!!!