Let $$C_1$$ be the circle in the third quadrant of radius 3, that touches both coordinate axes. Let $$C_2$$ be the circle with centre (1, 3) that touches $$C_1$$ externally at the point $$(\alpha, \beta)$$. If $$(\beta - \alpha)^2 = \frac{m}{n}$$, gcd(m, n) = 1, then $$m + n$$ is equal to :

Circle $$C_1$$ is in the third quadrant with radius 3, touching both coordinate axes, so its centre is $$(-3, -3)$$.

Circle $$C_2$$ has centre $$(1, 3)$$ and touches $$C_1$$ externally. The distance between centres is:

$$d = \sqrt{(1-(-3))^2 + (3-(-3))^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$$

Since they touch externally, $$r_1 + r_2 = d$$, so $$r_2 = 2\sqrt{13} - 3$$.

The point of external tangency $$(\alpha, \beta)$$ divides the line segment joining the centres internally in the ratio $$r_1 : r_2 = 3 : (2\sqrt{13} - 3)$$.

$$\alpha = \frac{3 \cdot 1 + (2\sqrt{13}-3)(-3)}{2\sqrt{13}} = \frac{3 - 6\sqrt{13} + 9}{2\sqrt{13}} = \frac{12 - 6\sqrt{13}}{2\sqrt{13}} = \frac{6}{\sqrt{13}} - 3$$

$$\beta = \frac{3 \cdot 3 + (2\sqrt{13}-3)(-3)}{2\sqrt{13}} = \frac{9 - 6\sqrt{13} + 9}{2\sqrt{13}} = \frac{18 - 6\sqrt{13}}{2\sqrt{13}} = \frac{9}{\sqrt{13}} - 3$$

$$\beta - \alpha = \frac{9}{\sqrt{13}} - \frac{6}{\sqrt{13}} = \frac{3}{\sqrt{13}}$$

$$(\beta - \alpha)^2 = \frac{9}{13}$$

So $$m = 9$$, $$n = 13$$, and $$\gcd(9, 13) = 1$$. Therefore $$m + n = 22$$.

Hence, the correct answer is Option C.