If $$\Delta_r = \begin{vmatrix} r & 2r-1 & 3r-2 \\ \frac{n}{2} & n-1 & a \\ \frac{1}{2}n(n-1) & (n-1)^2 & \frac{1}{2}(n-1)(3n+4) \end{vmatrix}$$, then the value of $$\sum_{r=1}^{n-1} \Delta_r$$:

We are given the determinant:

$$\Delta_r = \begin{vmatrix} r & 2r-1 & 3r-2 \\ \frac{n}{2} & n-1 & a \\ \frac{1}{2}n(n-1) & (n-1)^2 & \frac{1}{2}(n-1)(3n+4) \end{vmatrix}$$

We need to find the sum $$ S = \sum_{r=1}^{n-1} \Delta_r $$.

First, we expand the determinant $$\Delta_r$$ along the first row:

$$\Delta_r = r \begin{vmatrix} n-1 & a \\ (n-1)^2 & \frac{1}{2}(n-1)(3n+4) \end{vmatrix} - (2r-1) \begin{vmatrix} \frac{n}{2} & a \\ \frac{1}{2}n(n-1) & \frac{1}{2}(n-1)(3n+4) \end{vmatrix} + (3r-2) \begin{vmatrix} \frac{n}{2} & n-1 \\ \frac{1}{2}n(n-1) & (n-1)^2 \end{vmatrix}$$

Now, we compute each of the 2x2 determinants.

For the first minor:

$$\begin{vmatrix} n-1 & a \\ (n-1)^2 & \frac{1}{2}(n-1)(3n+4) \end{vmatrix} = (n-1) \cdot \frac{1}{2}(n-1)(3n+4) - a \cdot (n-1)^2 = (n-1)^2 \left( \frac{3n+4}{2} - a \right)$$

For the second minor:

$$\begin{vmatrix} \frac{n}{2} & a \\ \frac{1}{2}n(n-1) & \frac{1}{2}(n-1)(3n+4) \end{vmatrix} = \left( \frac{n}{2} \right) \cdot \left( \frac{1}{2}(n-1)(3n+4) \right) - a \cdot \left( \frac{1}{2}n(n-1) \right) = \frac{n(n-1)}{4} (3n+4) - \frac{a n (n-1)}{2} = \frac{n(n-1)}{4} \left( 3n+4 - 2a \right)$$

For the third minor:

$$\begin{vmatrix} \frac{n}{2} & n-1 \\ \frac{1}{2}n(n-1) & (n-1)^2 \end{vmatrix} = \left( \frac{n}{2} \right) \cdot (n-1)^2 - (n-1) \cdot \left( \frac{1}{2}n(n-1) \right) = \frac{n}{2} (n-1)^2 - \frac{n}{2} (n-1)^2 = 0$$

Substituting these minors back into the expression for $$\Delta_r$$:

$$\Delta_r = r \cdot \left[ (n-1)^2 \left( \frac{3n+4}{2} - a \right) \right] - (2r-1) \cdot \left[ \frac{n(n-1)}{4} (3n+4 - 2a) \right] + (3r-2) \cdot 0$$

Simplify the expression. Let $$ K = 3n + 4 - 2a $$. Note that:

$$\frac{3n+4}{2} - a = \frac{3n+4 - 2a}{2} = \frac{K}{2}$$

So,

$$\Delta_r = r \cdot (n-1)^2 \cdot \frac{K}{2} - (2r-1) \cdot \frac{n(n-1)}{4} K$$

Factor out $$ K $$:

$$\Delta_r = K \left[ \frac{r (n-1)^2}{2} - \frac{(2r-1) n (n-1)}{4} \right]$$

Distribute the terms inside the brackets:

$$\Delta_r = K \left[ \frac{r (n-1)^2}{2} - \frac{2r n (n-1)}{4} + \frac{n (n-1)}{4} \right] = K \left[ \frac{r (n-1)^2}{2} - \frac{r n (n-1)}{2} + \frac{n (n-1)}{4} \right]$$

Factor $$ (n-1) $$ from the first two terms:

$$\Delta_r = K \left[ r (n-1) \left( \frac{n-1}{2} - \frac{n}{2} \right) + \frac{n (n-1)}{4} \right] = K \left[ r (n-1) \left( \frac{-1}{2} \right) + \frac{n (n-1)}{4} \right] = K (n-1) \left( -\frac{r}{2} + \frac{n}{4} \right)$$

Simplify:

$$\Delta_r = K (n-1) \cdot \frac{1}{4} (n - 2r) = \frac{1}{4} (3n+4-2a) (n-1) (n - 2r)$$

Now, sum $$\Delta_r$$ from $$ r = 1 $$ to $$ r = n-1 $$:

$$S = \sum_{r=1}^{n-1} \Delta_r = \sum_{r=1}^{n-1} \frac{1}{4} (3n+4-2a) (n-1) (n - 2r)$$

Factor out the constants:

$$S = \frac{1}{4} (3n+4-2a) (n-1) \sum_{r=1}^{n-1} (n - 2r)$$

Compute the sum:

$$\sum_{r=1}^{n-1} (n - 2r) = \sum_{r=1}^{n-1} n - 2 \sum_{r=1}^{n-1} r = n(n-1) - 2 \cdot \frac{(n-1)n}{2} = n(n-1) - n(n-1) = 0$$

Therefore,

$$S = \frac{1}{4} (3n+4-2a) (n-1) \cdot 0 = 0$$

The sum $$ S = 0 $$, which is a constant independent of both $$ a $$ and $$ n $$.

Hence, the correct answer is Option A.