Let a line passing through the point $$(-1, 2, 3)$$ intersect the lines $$L_1: \frac{x-1}{3} = \frac{y-2}{2} = \frac{z+1}{-2}$$ at $$M(\alpha, \beta, \gamma)$$ and $$L_2: \frac{x+2}{-3} = \frac{y-2}{-2} = \frac{z-1}{4}$$ at $$N(a, b, c)$$. Then the value of $$\frac{(\alpha + \beta + \gamma)^2}{(a + b + c)^2}$$ equals ________________.

Line through $$(-1,2,3)$$ intersects $$L_1: \frac{x-1}{3}=\frac{y-2}{2}=\frac{z+1}{-2}$$ at $$M$$ and $$L_2: \frac{x+2}{-3}=\frac{y-2}{-2}=\frac{z-1}{4}$$ at $$N$$.

$$M = (3s+1, 2s+2, -2s-1)$$ on $$L_1$$. $$N = (-3t-2, -2t+2, 4t+1)$$ on $$L_2$$.

Direction from $$(-1,2,3)$$ to $$M$$: $$(3s+2, 2s, -2s-4)$$.

Direction from $$(-1,2,3)$$ to $$N$$: $$(-3t-1, -2t, 4t-2)$$.

These must be proportional: $$\frac{3s+2}{-3t-1} = \frac{2s}{-2t} = \frac{-2s-4}{4t-2}$$.

From $$\frac{2s}{-2t} = \frac{s}{-t}$$: Let ratio = $$k = s/(-t)$$, i.e., $$s = -kt$$.

From $$\frac{3s+2}{-3t-1} = \frac{s}{-t}$$: $$-t(3s+2) = s(-3t-1) \Rightarrow -3st-2t = -3st-s \Rightarrow -2t = -s \Rightarrow s = 2t$$.

From $$\frac{-2s-4}{4t-2} = \frac{s}{-t}$$: $$-t(-2s-4) = s(4t-2) \Rightarrow 2st+4t = 4st-2s \Rightarrow 4t+2s = 2st$$.

With $$s = 2t$$: $$4t + 4t = 4t^2 \Rightarrow 8t = 4t^2 \Rightarrow t = 2$$, $$s = 4$$.

$$M = (13, 10, -9)$$: $$\alpha+\beta+\gamma = 14$$.

$$N = (-8, -2, 9)$$: $$a+b+c = -1$$.

$$\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2} = \frac{196}{1} = 196$$.

Therefore, the answer is $$\boxed{196}$$.