The positive value of the determinant of the matrix $$A$$, whose $$Adj(Adj(A)) = \begin{bmatrix} 14 & 28 & -14 \\ -14 & 14 & 28 \\ 28 & -14 & 14 \end{bmatrix}$$, is ______

Using the identity for adjugate of adjugate of an $$n\times n$$ matrix $$A$$, namely $$\text{Adj}(\text{Adj}(A)) = |A|^{n-2}\cdot A$$, and substituting $$n=3$$ gives $$\text{Adj}(\text{Adj}(A)) = |A|\cdot A$$.

Therefore, we have $$|A|\cdot A = B$$ where $$B$$ is the given matrix. Taking determinants on both sides yields $$|A|^3\cdot|A| = |B|$$, hence $$|A|^4 = |B|$$.

To compute $$|B|$$, note that

$$B = \begin{bmatrix}14 & 28 & -14 \\ -14 & 14 & 28 \\ 28 & -14 & 14\end{bmatrix} = 14\begin{bmatrix}1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1\end{bmatrix}$$

so

$$|B| = 14^3 \cdot \begin{vmatrix}1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1\end{vmatrix}$$

Expanding this determinant:

$$= 1(1\cdot1 - 2\cdot(-1)) - 2((-1)(1) - 2\cdot2) + (-1)((-1)(-1) - 1\cdot2)$$

$$= 1(1+2) - 2(-1-4) + (-1)(1-2)$$

$$= 3 - 2(-5) + (-1)(-1)$$

$$= 3 + 10 + 1 = 14$$

Hence

$$|B| = 14^3 \times 14 = 14^4$$

Substituting back into $$|A|^4 = |B|$$ gives $$|A|^4 = 14^4$$ so $$|A| = \pm14$$. The positive value is $$|A| = 14$$.

The answer is $$\boxed{14}$$.