The base of an equilateral triangle is along the line given by $$3x + 4y = 9$$. If a vertex of the triangle is $$(1, 2)$$, then the length of a side of the triangle is:

The base of the equilateral triangle lies along the line $$3x + 4y = 9$$, and one vertex is at $$(1, 2)$$. Since it is an equilateral triangle, the altitude from the vertex $$(1, 2)$$ to the base will be perpendicular to the base and will bisect the base. This altitude can be used to find the side length.

First, find the slope of the base line $$3x + 4y = 9$$. Rewrite the equation in slope-intercept form:

$$4y = -3x + 9$$

$$y = -\frac{3}{4}x + \frac{9}{4}$$

The slope of the base is $$-\frac{3}{4}$$. The slope of the altitude, which is perpendicular to the base, is the negative reciprocal. So, the slope of the altitude is:

$$m = -\frac{1}{-\frac{3}{4}} = \frac{4}{3}$$

The altitude passes through the vertex $$(1, 2)$$, so its equation is:

$$y - 2 = \frac{4}{3}(x - 1)$$

Simplify this equation:

$$y - 2 = \frac{4}{3}x - \frac{4}{3}$$

$$y = \frac{4}{3}x - \frac{4}{3} + 2$$

$$y = \frac{4}{3}x - \frac{4}{3} + \frac{6}{3}$$

$$y = \frac{4}{3}x + \frac{2}{3}$$

The foot of the perpendicular (denoted as point D) is the intersection of the altitude and the base line. Solve the system of equations:

Base: $$3x + 4y = 9$$

Altitude: $$y = \frac{4}{3}x + \frac{2}{3}$$

Substitute $$y$$ from the altitude equation into the base equation:

$$3x + 4\left(\frac{4}{3}x + \frac{2}{3}\right) = 9$$

$$3x + \frac{16}{3}x + \frac{8}{3} = 9$$

Multiply through by 3 to clear denominators:

$$9x + 16x + 8 = 27$$

$$25x + 8 = 27$$

$$25x = 19$$

$$x = \frac{19}{25}$$

Now substitute $$x = \frac{19}{25}$$ into the altitude equation to find $$y$$:

$$y = \frac{4}{3} \cdot \frac{19}{25} + \frac{2}{3}$$

$$y = \frac{76}{75} + \frac{2}{3}$$

Convert $$\frac{2}{3}$$ to a fraction with denominator 75:

$$\frac{2}{3} = \frac{2 \times 25}{3 \times 25} = \frac{50}{75}$$

$$y = \frac{76}{75} + \frac{50}{75} = \frac{126}{75}$$

Simplify $$\frac{126}{75}$$ by dividing numerator and denominator by 3:

$$y = \frac{42}{25}$$

So, point D is at $$\left(\frac{19}{25}, \frac{42}{25}\right)$$.

The distance between vertex A $$(1, 2)$$ and point D $$\left(\frac{19}{25}, \frac{42}{25}\right)$$ is the length of the altitude. Compute the differences:

$$\Delta x = \frac{19}{25} - 1 = \frac{19}{25} - \frac{25}{25} = -\frac{6}{25}$$

$$\Delta y = \frac{42}{25} - 2 = \frac{42}{25} - \frac{50}{25} = -\frac{8}{25}$$

The distance AD is:

$$\text{AD} = \sqrt{\left(-\frac{6}{25}\right)^2 + \left(-\frac{8}{25}\right)^2} = \sqrt{\frac{36}{625} + \frac{64}{625}} = \sqrt{\frac{100}{625}} = \frac{\sqrt{100}}{\sqrt{625}} = \frac{10}{25} = \frac{2}{5}$$

In an equilateral triangle with side length $$s$$, the height $$h$$ is given by $$h = \frac{\sqrt{3}}{2} s$$. Here, AD is the height, so:

$$\frac{2}{5} = \frac{\sqrt{3}}{2} s$$

Solve for $$s$$:

$$s = \frac{2}{5} \times \frac{2}{\sqrt{3}} = \frac{4}{5\sqrt{3}}$$

Rationalize the denominator:

$$s = \frac{4}{5\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3}}{15}$$

Thus, the length of a side of the triangle is $$\frac{4\sqrt{3}}{15}$$. Comparing with the options, this matches option B.

Hence, the correct answer is Option B.