Let $$A = \begin{bmatrix} 2 & 3 \\ a & 0 \end{bmatrix}$$, $$a \in R$$ be written as $$P + Q$$ where $$P$$ is a symmetric matrix and $$Q$$ is skew symmetric matrix. If det$$(Q) = 9$$, then the modulus of the sum of all possible values of determinant of $$P$$ is equal to:

Any matrix $$A$$ can be written as $$P + Q$$ where $$P = \frac{A + A^T}{2}$$ (symmetric) and $$Q = \frac{A - A^T}{2}$$ (skew-symmetric).

With $$A = \begin{bmatrix}2 & 3 \\ a & 0\end{bmatrix}$$, we get: $$P = \begin{bmatrix}2 & \frac{3+a}{2} \\ \frac{3+a}{2} & 0\end{bmatrix}, \quad Q = \begin{bmatrix}0 & \frac{3-a}{2} \\ \frac{a-3}{2} & 0\end{bmatrix}.$$

Computing $$\det(Q)$$: $$\det(Q) = 0 \cdot 0 - \frac{3-a}{2} \cdot \frac{a-3}{2} = \frac{(3-a)^2}{4} = 9 \implies (3-a)^2 = 36 \implies 3-a = \pm 6.$$

So $$a = -3$$ or $$a = 9$$.

Computing $$\det(P) = 2 \cdot 0 - \left(\frac{3+a}{2}\right)^2 = -\frac{(3+a)^2}{4}$$.

For $$a = -3$$: $$\det(P) = -\frac{0}{4} = 0$$.

For $$a = 9$$: $$\det(P) = -\frac{144}{4} = -36$$.

The sum of all possible values of $$\det(P)$$ is $$0 + (-36) = -36$$. The modulus of this sum is $$\boxed{36}$$.