Let $$P = \begin{pmatrix} 3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0 \end{pmatrix}$$, where $$\alpha \in R$$. Suppose $$Q = [q_{ij}]$$ is a matrix satisfying $$PQ = kI_3$$ for some non-zero $$k \in R$$. If $$q_{23} = -\frac{k}{8}$$ and $$Q = \frac{k^2}{2}$$, then $$\alpha^2 + k^2$$ is equal to ______.

We are given $$P = \begin{pmatrix} 3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0 \end{pmatrix}$$ and $$PQ = kI_3$$ for some non-zero $$k$$. This means $$Q = kP^{-1} = \frac{k}{\det(P)} \text{adj}(P)$$.

We first find $$\det(P)$$. Expanding along the first row, $$\det(P) = 3(0 \cdot 0 - \alpha \cdot (-5)) - (-1)(2 \cdot 0 - \alpha \cdot 3) + (-2)(2 \cdot (-5) - 0 \cdot 3)$$.

$$= 3(5\alpha) + 1(-3\alpha) + (-2)(-10) = 15\alpha - 3\alpha + 20 = 12\alpha + 20$$.

Now we use the condition $$q_{23} = -\frac{k}{8}$$. Since $$Q = \frac{k}{\det(P)} \text{adj}(P)$$, we have $$q_{23} = \frac{k}{\det(P)} \cdot C_{32}$$, where $$C_{32}$$ is the cofactor of the $$(3, 2)$$ entry of $$P$$.

$$C_{32} = (-1)^{3+2} \begin{vmatrix} 3 & -2 \\ 2 & \alpha \end{vmatrix} = -(3\alpha + 4)$$.

So $$q_{23} = \frac{k \cdot (-(3\alpha + 4))}{12\alpha + 20} = -\frac{k}{8}$$.

This gives $$\frac{3\alpha + 4}{12\alpha + 20} = \frac{1}{8}$$. Cross-multiplying, $$8(3\alpha + 4) = 12\alpha + 20$$, so $$24\alpha + 32 = 12\alpha + 20$$, giving $$12\alpha = -12$$ and $$\alpha = -1$$.

Substituting $$\alpha = -1$$, we get $$\det(P) = 12(-1) + 20 = 8$$.

Now we use the condition $$|Q| = \frac{k^2}{2}$$. From $$PQ = kI$$, taking determinants, $$\det(P) \cdot \det(Q) = k^3$$, so $$\det(Q) = \frac{k^3}{\det(P)} = \frac{k^3}{8}$$.

Setting this equal to $$\frac{k^2}{2}$$, we get $$\frac{k^3}{8} = \frac{k^2}{2}$$. Since $$k \neq 0$$, dividing both sides by $$k^2$$ gives $$\frac{k}{8} = \frac{1}{2}$$, so $$k = 4$$.

Therefore, $$\alpha^2 + k^2 = (-1)^2 + 4^2 = 1 + 16 = 17$$.

Hence, the answer is $$17$$.