The distance of the origin from the centroid of the triangle whose two sides have the equations $$x - 2y + 1 = 0$$ and $$2x - y - 1 = 0$$ and whose orthocenter is $$\left(\frac{7}{3}, \frac{7}{3}\right)$$ is:

We are given two sides of a triangle with equations $$x - 2y + 1 = 0$$ and $$2x - y - 1 = 0$$, and the orthocenter is $$H = \left(\frac{7}{3}, \frac{7}{3}\right)$$. Solving $$x - 2y + 1 = 0$$ and $$2x - y - 1 = 0$$ simultaneously, from the first equation we get $$x = 2y - 1$$. Substituting into the second equation gives $$2(2y-1) - y - 1 = 0 \Rightarrow 3y - 3 = 0 \Rightarrow y = 1$$, so $$x = 1$$, giving vertex $$C = (1, 1)$$.

Let the triangle have vertices $$A$$ on line $$2x - y - 1 = 0$$, $$B$$ on line $$x - 2y + 1 = 0$$, and $$C = (1, 1)$$. Side $$CA$$ lies on $$2x - y - 1 = 0$$ (slope = 2) and side $$CB$$ lies on $$x - 2y + 1 = 0$$ (slope = 1/2). The altitude from $$B$$ is perpendicular to $$CA$$ (slope 2), so its slope is $$-1/2$$, and the altitude from $$A$$ is perpendicular to $$CB$$ (slope 1/2), so its slope is $$-2$$.

Altitude from $$B$$ passes through $$H = (7/3, 7/3)$$ with slope $$-1/2$$, so $$y - 7/3 = -1/2(x - 7/3)$$, which simplifies to $$y = -x/2 + 7/6 + 7/3 = -x/2 + 7/2$$. Since $$B$$ lies on $$x - 2y + 1 = 0$$, i.e., $$x = 2y - 1$$, we have $$y = -(2y-1)/2 + 7/2 = -y + 1/2 + 7/2$$, giving $$2y = 4$$, so $$y = 2$$ and $$x = 3$$. Thus $$B = (3, 2)$$.

Similarly, the altitude from $$A$$ passes through $$H = (7/3, 7/3)$$ with slope $$-2$$, so $$y - 7/3 = -2(x - 7/3)$$, leading to $$y = -2x + 14/3 + 7/3 = -2x + 7$$. Since $$A$$ lies on $$2x - y - 1 = 0$$, i.e., $$y = 2x - 1$$, we set $$2x - 1 = -2x + 7$$, yielding $$4x = 8$$, hence $$x = 2$$ and $$y = 3$$, and so $$A = (2, 3)$$.

The centroid is $$G = \left(\frac{1+3+2}{3}, \frac{1+2+3}{3}\right) = \left(\frac{6}{3}, \frac{6}{3}\right) = (2, 2)$$, and the distance from the origin to the centroid is $$d = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$$. Therefore, the answer is Option C: $$\boldsymbol{2\sqrt{2}}$$.