If $$A(3, 1, -1)$$, $$B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right)$$, $$C(2, 2, 1)$$ and $$D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)$$ are the vertices of a quadrilateral $$ABCD$$, then its area is

We are given $$A(3, 1, -1)$$, $$B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right)$$, $$C(2, 2, 1)$$, $$D\left(\frac{10}{3}, \frac{2}{3}, -\frac{1}{3}\right)$$.

First, let us check whether $$ABCD$$ is a parallelogram by computing the side vectors:

$$\overrightarrow{AB} = B - A = \left(\frac{5}{3} - 3, \frac{7}{3} - 1, \frac{1}{3} + 1\right) = \left(-\frac{4}{3}, \frac{4}{3}, \frac{4}{3}\right)$$

$$\overrightarrow{DC} = C - D = \left(2 - \frac{10}{3}, 2 - \frac{2}{3}, 1 + \frac{1}{3}\right) = \left(-\frac{4}{3}, \frac{4}{3}, \frac{4}{3}\right)$$

Since $$\overrightarrow{AB} = \overrightarrow{DC}$$, the quadrilateral $$ABCD$$ is a parallelogram.

For a parallelogram with adjacent sides $$\overrightarrow{AB}$$ and $$\overrightarrow{AD}$$, the area is $$|\overrightarrow{AB} \times \overrightarrow{AD}|$$.

Computing $$\overrightarrow{AD}$$:

$$\overrightarrow{AD} = D - A = \left(\frac{10}{3} - 3, \frac{2}{3} - 1, -\frac{1}{3} + 1\right) = \left(\frac{1}{3}, -\frac{1}{3}, \frac{2}{3}\right)$$

Computing the cross product $$\overrightarrow{AB} \times \overrightarrow{AD}$$:

$$\overrightarrow{AB} \times \overrightarrow{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -\frac{4}{3} & \frac{4}{3} & \frac{4}{3} \\ \frac{1}{3} & -\frac{1}{3} & \frac{2}{3} \end{vmatrix}$$

$$\hat{i}$$: $$\frac{4}{3} \cdot \frac{2}{3} - \frac{4}{3} \cdot \left(-\frac{1}{3}\right) = \frac{8}{9} + \frac{4}{9} = \frac{12}{9} = \frac{4}{3}$$

$$\hat{j}$$: $$-\left[\left(-\frac{4}{3}\right)\left(\frac{2}{3}\right) - \frac{4}{3} \cdot \frac{1}{3}\right] = -\left[-\frac{8}{9} - \frac{4}{9}\right] = -\left(-\frac{12}{9}\right) = \frac{4}{3}$$

$$\hat{k}$$: $$\left(-\frac{4}{3}\right)\left(-\frac{1}{3}\right) - \frac{4}{3} \cdot \frac{1}{3} = \frac{4}{9} - \frac{4}{9} = 0$$

So $$\overrightarrow{AB} \times \overrightarrow{AD} = \left(\frac{4}{3}, \frac{4}{3}, 0\right)$$.

$$|\overrightarrow{AB} \times \overrightarrow{AD}| = \sqrt{\left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^2 + 0} = \frac{4}{3}\sqrt{2}$$

Therefore, the area of the parallelogram $$ABCD$$ is:

$$\text{Area} = \frac{4\sqrt{2}}{3}$$

The answer is Option D: $$\frac{4\sqrt{2}}{3}$$.