One of the points of intersection of the curves $$y = 1 + 3x - 2x^2$$ and $$y = \frac{1}{x}$$ is $$\left(\frac{1}{2}, 2\right)$$. Let the area of the region enclosed by these curves be $$\frac{1}{24}(l\sqrt{5}+m) - n\log_e(1+\sqrt{5})$$, where $$l, m, n \in N$$. Then $$l + m + n$$ is equal to:

We need to find the area enclosed by the curves $$y = 1 + 3x - 2x^2$$ and $$y = \frac{1}{x}$$, given that one intersection point is $$\left(\frac{1}{2}, 2\right)$$.

At intersection: $$1 + 3x - 2x^2 = \frac{1}{x}$$

$$x(1 + 3x - 2x^2) = 1$$

$$x + 3x^2 - 2x^3 = 1$$

$$2x^3 - 3x^2 - x + 1 = 0$$

We know $$x = \frac{1}{2}$$ is a root. Factor out $$(2x - 1)$$:

$$2x^3 - 3x^2 - x + 1 = (2x - 1)(x^2 - x - 1) = 0$$

From $$x^2 - x - 1 = 0$$: $$x = \frac{1 \pm \sqrt{5}}{2}$$

So the roots are $$x = \frac{1}{2}$$, $$x = \frac{1 + \sqrt{5}}{2}$$ (the golden ratio $$\phi$$), and $$x = \frac{1 - \sqrt{5}}{2}$$ (negative, so we consider the region for $$x > 0$$).

For $$x$$ between $$\frac{1}{2}$$ and $$\frac{1+\sqrt{5}}{2}$$, test at $$x = 1$$:

Parabola: $$1 + 3 - 2 = 2$$

Hyperbola: $$\frac{1}{1} = 1$$

So the parabola is above the hyperbola in this interval.

$$A = \int_{1/2}^{(1+\sqrt{5})/2} \left(1 + 3x - 2x^2 - \frac{1}{x}\right) dx$$

$$= \left[x + \frac{3x^2}{2} - \frac{2x^3}{3} - \ln x\right]_{1/2}^{(1+\sqrt{5})/2}$$

Let $$\phi = \frac{1+\sqrt{5}}{2}$$. Note that $$\phi^2 = \phi + 1 = \frac{3+\sqrt{5}}{2}$$ and $$\phi^3 = 2\phi + 1 = 2 + \sqrt{5}$$.

At $$x = \phi$$:

$$\phi + \frac{3\phi^2}{2} - \frac{2\phi^3}{3} - \ln\phi$$

$$= \frac{1+\sqrt{5}}{2} + \frac{3(3+\sqrt{5})}{4} - \frac{2(2+\sqrt{5})}{3} - \ln\phi$$

$$= \frac{1+\sqrt{5}}{2} + \frac{9+3\sqrt{5}}{4} - \frac{4+2\sqrt{5}}{3} - \ln\phi$$

$$= \frac{6(1+\sqrt{5}) + 3(9+3\sqrt{5}) - 4(4+2\sqrt{5})}{12} - \ln\phi$$

$$= \frac{6+6\sqrt{5}+27+9\sqrt{5}-16-8\sqrt{5}}{12} - \ln\phi$$

$$= \frac{17+7\sqrt{5}}{12} - \ln\phi$$

At $$x = \frac{1}{2}$$:

$$\frac{1}{2} + \frac{3}{8} - \frac{1}{12} - \ln\frac{1}{2} = \frac{12+9-2}{24} + \ln 2 = \frac{19}{24} + \ln 2$$

Area:

$$A = \frac{17+7\sqrt{5}}{12} - \ln\phi - \frac{19}{24} - \ln 2$$

$$= \frac{34+14\sqrt{5}-19}{24} - \ln(2\phi)$$

$$= \frac{15+14\sqrt{5}}{24} - \ln(1+\sqrt{5})$$

This can be written as $$\frac{1}{24}(14\sqrt{5} + 15) - \ln(1+\sqrt{5})$$.

Comparing with $$\frac{1}{24}(l\sqrt{5} + m) - n\log_e(1+\sqrt{5})$$:

$$l = 14, m = 15, n = 1$$

$$l + m + n = 14 + 15 + 1 = 30$$

The correct answer is Option (3): $$\boxed{30}$$.