Divide both sides by $abc$. The equation reads $4=\left( 1+\frac{3}{a} \right) \left(1+\frac{3}{b} \right) \left(1+\frac{3}{c} \right)$. If none of $a,b,c=4$, the greatest possible value of $\left( 1+\frac{3}{a} \right) \left(1+\frac{3}{b} \right) \left(1+\frac{3}{c} \right)$ with $a$, $b$, and $c$ not all equal would be $\frac{8}{5} \cdot \frac{8}{5} \cdot \frac{9}{6} < 4$, therefore at least one of them is equal to $4$; WLOG it's $a$. Then we have $\frac{16}{7}=\left(1+\frac{3}{b} \right) \left(1+\frac{3}{c} \right)$ implying that at least one of $b$ and $c$ is divisible by $7$ and is probably equal to $7$ (anything larger would make the product too small); WLOG $b=7$. By substituting this into the previous equation we get, $c=5$ so $a+b+c=4+7+5=16$

First, notice that we can't have $a, b, c \geq 6$, since otherwise \[(a + 3)(b + 3)(c + 3) \leq \left(\frac32a\right) \left(\frac32b\right) \left(\frac32c\right) = \frac{27}{8}abc < 4abc.\]So without loss of generality suppose $a \in \{5, 6\}$. If we set $a = 5$, the equation becomes $20bc = 8(b + 3)(c + 3)$, or $bc - 2b - 2c - 6 = 0$. This rewrites as $(b - 2)(c - 2) = 10$, which gives $(b, c) = (4, 7)$ or vice versa. Similarly, $a = 6$ yields \[24bc = 9(b + 3)(c + 3) \implies 25bc - 45b - 45c - 135 = 0\]\[ \implies (5b - 9)(5c - 9) = 216,\]which gives $(b, c) = (2, 45)$ and $(b, c) = (3, 9)$. These are both too small, so the only solutions are $(a, b, c) = (4, 5, 7)$ and permutations. Thus the answer is $4 + 5 + 7 = \fbox{16}$.