$$ \text{Let } A \text{ be a } 3\times 3 \text{ matrix such that } X^TAX=0 \text{ for all nonzero } 3\times1 \text{ matrices } X=\begin{bmatrix}x\\y\\z\end{bmatrix}. \text{ If } A\begin{bmatrix}1\\1\\1\end{bmatrix} = \begin{bmatrix}1\\4\\-5\end{bmatrix}, \; A\begin{bmatrix}1\\2\\1\end{bmatrix} = \begin{bmatrix}0\\4\\-8\end{bmatrix}, \text{ and } \det(\operatorname{adj}(2(A+I)))=2^\alpha 3^\beta 5^\gamma, \; \alpha,\beta,\gamma\in\mathbb{N}, \text{ then } \alpha^2+\beta^2+\gamma^2 \text{ is:} \underline{\hspace{2cm}}$$

Since $$X^TAX = 0$$ for all nonzero $$3 \times 1$$ matrices $$X$$, the matrix $$A$$ must be skew-symmetric: $$A^T = -A$$. A $$3 \times 3$$ skew-symmetric matrix has the form $$A = \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix}$$.

From $$A\begin{pmatrix}1\\1\\1\end{pmatrix} = \begin{pmatrix}1\\4\\-5\end{pmatrix}$$, we get $$a + b = 1$$, $$-a + c = 4$$, and $$-b - c = -5$$. From $$A\begin{pmatrix}1\\2\\1\end{pmatrix} = \begin{pmatrix}0\\4\\-8\end{pmatrix}$$, the first component gives $$2a + b = 0$$, so $$a = -1$$, $$b = 2$$, $$c = 3$$.

Now $$A + I = \begin{pmatrix} 1 & -1 & 2 \\ 1 & 1 & 3 \\ -2 & -3 & 1 \end{pmatrix}$$ and $$2(A+I) = \begin{pmatrix} 2 & -2 & 4 \\ 2 & 2 & 6 \\ -4 & -6 & 2 \end{pmatrix}$$. Computing $$\det(2(A+I)) = 2(4+36) + 2(4+24) + 4(-12+8) = 80 + 56 - 16 = 120$$. For a $$3 \times 3$$ matrix, $$\det(\text{adj}(B)) = (\det B)^{n-1} = 120^2 = 14400 = 2^6 \cdot 3^2 \cdot 5^2$$.

Thus $$\alpha = 6$$, $$\beta = 2$$, $$\gamma = 2$$, and $$\alpha^2 + \beta^2 + \gamma^2 = 36 + 4 + 4 = \boxed{44}$$.