If $$A(1, -1, 2), B(5, 7, -6), C(3, 4, -10)$$ and $$D(-1, -4, -2)$$ are the vertices of a quadrilateral $$ABCD$$, then its area is :

We are given $$A(1, -1, 2)$$, $$B(5, 7, -6)$$, $$C(3, 4, -10)$$, $$D(-1, -4, -2)$$.

First, let us check whether $$ABCD$$ is a parallelogram by computing the midpoints of the diagonals.

Midpoint of $$AC = \left(\frac{1+3}{2}, \frac{-1+4}{2}, \frac{2+(-10)}{2}\right) = (2, 1.5, -4)$$

Midpoint of $$BD = \left(\frac{5+(-1)}{2}, \frac{7+(-4)}{2}, \frac{-6+(-2)}{2}\right) = (2, 1.5, -4)$$

Since both diagonals have the same midpoint, $$ABCD$$ is a parallelogram.

For a parallelogram, the area can be computed using the diagonal vectors with the formula:

$$\text{Area} = \frac{1}{2}|\overrightarrow{AC} \times \overrightarrow{BD}|$$

Computing the diagonal vectors:

$$\overrightarrow{AC} = C - A = (3-1, 4-(-1), -10-2) = (2, 5, -12)$$

$$\overrightarrow{BD} = D - B = (-1-5, -4-7, -2-(-6)) = (-6, -11, 4)$$

Now compute the cross product $$\overrightarrow{AC} \times \overrightarrow{BD}$$:

$$\overrightarrow{AC} \times \overrightarrow{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 5 & -12 \\ -6 & -11 & 4 \end{vmatrix}$$

$$\hat{i}$$-component: $$5 \times 4 - (-12) \times (-11) = 20 - 132 = -112$$

$$\hat{j}$$-component: $$-(2 \times 4 - (-12) \times (-6)) = -(8 - 72) = 64$$

$$\hat{k}$$-component: $$2 \times (-11) - 5 \times (-6) = -22 + 30 = 8$$

So $$\overrightarrow{AC} \times \overrightarrow{BD} = (-112, 64, 8)$$.

$$|\overrightarrow{AC} \times \overrightarrow{BD}| = \sqrt{(-112)^2 + 64^2 + 8^2}$$

$$= \sqrt{12544 + 4096 + 64}$$

$$= \sqrt{16704}$$

Let us simplify $$\sqrt{16704}$$. We find: $$16704 = 576 \times 29$$, since $$576 = 24^2$$ and $$576 \times 29 = 16704$$.

$$\sqrt{16704} = 24\sqrt{29}$$

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

$$\text{Area} = \frac{1}{2} \times 24\sqrt{29} = 12\sqrt{29}$$

The answer is Option B: $$12\sqrt{29}$$.