Solution:

Solution 1: Notice that any of $a$, $b$, or $c$ being $3$ or $-2$ implies $a = b = c$, which is invalid. Thus,
$$(a^{2} - 9)(b^{2} - 9)(c^{2} - 9) = (b - 3)(c - 3)(a - 3)\implies (a + 3)(b + 3)(c + 3) = 1,$$
$$(a^{2} - 4)(b^{2} - 4)(c^{2} - 4) = (b + 2)(c + 2)(a + 2)\implies (a - 2)(b - 2)(c - 2) = 1.$$
Therefore, $2$ and $-3$ are roots of the polynomial $(x - a)(x - b)(x - c) + 1$, and so there exists some $t$ such that
$$(x - t)(x - 2)(x + 3) = (x - a)(x - b)(x - c) + 1.$$
Comparing coefficients gives $a + b + c = t - 1$ and $ab + bc + ca = - (t + 6)$. We can then solve for $t$ by noting $a^{2} + b^{2} + c^{2} = (b + 6) + (c + 6) + (a + 6) = a + b + c + 18$, so
$$ab + bc + ca = \frac{1}{2} ((a + b + c)^{2} - (a^{2} + b^{2} + c^{2})) = \frac{1}{2} ((a + b + c)^{2} - (a + b + c + 18)).$$
Hence,
$$- (t + 6) = \frac{1}{2} ((t - 1)^{2} - (t + 17))\Longrightarrow t^{2} - t - 4 = 0\Longrightarrow t = \frac{1\pm\sqrt{17}}{2}.$$
Therefore $a + b + c = \boxed{\frac{- 1\pm\sqrt{17}}{2}}$ are the two possible values of $a + b + c$.

Solution 2: Let $s = a + b + c$. Subtracting two adjacent equations gives $a^{2} - b^{2} = b - c$, or $(a - b)(a + b) = (b - c)$. Multiplying this and its cyclic variants gives
$$(a + b)(b + c)(c + a) = 1.$$
Now, we recall the identity
$$(a + b + c)^{3} = a^{3} + b^{3} + c^{3} + 3(a + b)(b + c)(c + a)$$
$$\qquad \Rightarrow \qquad s^{3} = a^{3} + b^{3} + c^{3} + 3.$$
To simplify $a^{3} + b^{3} + c^{3}$, we add $a$ times the first equation, $b$ times the second, and $c$ times the third to obtain
$$a^{3} + b^{3} + c^{3} = a(b + 6) + b(c + 6) + c(a + 6)$$
$$\qquad = (ab + bc + ca) + 6s$$
$$\qquad = \frac{1}{2}\Big((a + b + c)^{2} - (a^{2} + b^{2} + c^{2})\Big) + 6s$$
$$\qquad = \frac{1}{2} s^{2} - \frac{1}{2}\Big((b + 6) + (c + 6) + (a + 6)\Big) + 6s$$
$$\qquad = \frac{1}{2} s^{2} + \frac{11}{2} s - 9.$$
Therefore,
$$s^{3} = \frac{1}{2} s^{2} + \frac{11}{2} s - 6 \Rightarrow \left(s - \frac{3}{2}\right)\left(s^{2} + s - 4\right) = 0.$$
At this point, the only reasonable guess is that $s = \frac{3}{2}$ is an extra solution, and the remaining two roots $s = \frac{- 1\pm\sqrt{17}}{2}$ are the possible answers. We now justify this guess. Assume for sake of contradiction that $s = \frac{3}{2}$. Then,
$$a^{2} + b^{2} + c^{2} = (b + 6) + (c + 6) + (a + 6) = \frac{39}{2}$$
$$ab + bc + ca = \frac{1}{2}\left(\frac{9}{4} -\frac{39}{2}\right) = -\frac{69}{8}.$$
Then, observe
$$abc = (a + b + c)(ab + bc + ca) - (a + b)(b + c)(c + a)$$
$$= -\frac{207}{16} -1 = -\frac{223}{16}.$$
On the other hand,
$$(a + 6)(b + 6)(c + 6) = 216 + 36(a + b + c) + 6(ab + bc + ca) + abc$$
$$\qquad = 216 + 36\cdot \frac{3}{2} -6\cdot \frac{69}{8} -\frac{223}{16},$$
which is a rational number of denominator $16$. But $(a + 6)(b + 6)(c + 6) = b^{2}c^{2}a^{2} = \left(-\frac{223}{16}\right)^{2}$ has denominator $16^{2} = 256$, a contradiction. Thus $s = \frac{3}{2}$ is impossible. (It arises from $a = b = c = \frac{1}{2}$ which satisfies $(a + b)(b + c)(c + a) = 1$ but not the given conditions.)

Solution 3: Subtracting any two adjacent equations gives $a^{2} - b^{2} = b - c$, which is equivalent to both $(a - b)(a + b) = (b - c)$ and $(a - b)(a + b + 1) = (a - c)$. Multiplying each of these with its respective cyclic variants and canceling the $(a - b)(b - c)(c - a)$ factor (which is given to be nonzero), we get
$$(a + b)(b + c)(c + a) = 1\quad \mathrm{and}\quad (a + b + 1)(b + c + 1)(c + a + 1) = -1.$$
Expanding the latter equation and using the given equations gives the following result.
$$(a + b)(b + c)(c + a) + (a^{2} + b^{2} + c^{2}) + 3(ab + bc + ca) + 2(a + b + c) + 1 = -1$$
$$1 + (b + 6 + c + 6 + a + 6) + 3(ab + bc + ca) + 2(a + b + c) + 1 = -1$$
$$3(a + b + c) + 3(ab + bc + ca) = -21$$
$$a + b + c + ab + bc + ca = -7.$$
Let $s = a + b + c$. We can then solve for $s$ by considering the following:
$$s^{2} = (a^{2} + b^{2} + c^{2}) + 2(ab + bc + ca)$$
$$\quad = (b + 6 + c + 6 + a + 6) + 2(-7 - a - b - c)$$
$$\quad = -s + 4,$$
so $s = \frac{- 1\pm\sqrt{17}}{2}$.

Solution 4: Let $s = a + b + c$ and consider the polynomial
$$x + (x^{2} - 6) + ((x^{2} - 6)^{2} - 6) - s = x^{4} - 11x^{2} + x + 24 - s.$$
This polynomial has roots $a$, $b$, and $c$. By Vieta's, the sum of all four roots is $0$, so its fourth root must be $-s$. Using Vieta's again, we have $ab + bc + ca - sa - sb - sc = -11$. We can now solve for $s$.
$$ab + bc + ca - (a + b + c)^{2} = -11$$
$$a^{2} + b^{2} + c^{2} + ab + bc + ca = 11$$
$$\frac{1}{2} ((a + b + c)^{2} + (a^{2} + b^{2} + c^{2})) = 11$$
$$(a + b + c)^{2} + (b + 6 + c + 6 + a + 6) = 22$$
$$s^{2} + s - 4 = 0\Longrightarrow s = \frac{-1\pm\sqrt{17}}{2}.$$
Remark. Another way to finish using this approach is to substitute $-s$ directly into $x^{4} - 11x^{2} + x + 24 - s = 0$ to get $(s - 3)(s + 2)(x^{2} + x - 4) = 0$, then discard the solutions $s = 3$ and $s = -2$, which arise from the invalid values $a = b = c = 3$ and $a = b = c = -2$. (In the invalid cases, $s \neq a + b + c$ because $a = b = c$ is only a single root to the polynomial.)