Consider the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ having one of its focus at P(-3, 0). If the latus rectum through its other focus subtends a right angle at P and $$a^2b^2 = \alpha\sqrt{2} - \beta$$, $$\alpha, \beta \in \mathbb{N}$$. 

The standard form of a hyperbola with centre at the origin and transverse axis along the x-axis is
$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \qquad\left(a\gt 0,\;b\gt 0\right).$$

Its foci are $$\left(\pm c,0\right)$$ where $$c^{2}=a^{2}+b^{2}$$.

One focus is given as $$P(-3,0)$$, hence
$$c = 3 \quad\text{and}\quad c^{2}=9.$$ Therefore
$$a^{2}+b^{2}=9 \;.(1)$$

The other focus is $$S(3,0)$$. Through this focus the latus-rectum is the line $$x=3$$ whose end-points are $$Q\!\left(3,\;\frac{b^{2}}{a}\right), \quad R\!\left(3,\;-\frac{b^{2}}{a}\right).$$ The segment $$QR$$ subtends a right angle at the point $$P(-3,0)$$, i.e.
$$\angle QPR = 90^{\circ}.$$

For perpendicular vectors their dot product is zero. Vector $$\overrightarrow{PQ} = (3-(-3),\,\tfrac{b^{2}}{a}-0) = (6,\tfrac{b^{2}}{a})$$
Vector $$\overrightarrow{PR} = (3-(-3),\,-\tfrac{b^{2}}{a}-0) = (6,\,-\tfrac{b^{2}}{a}).$$

Dot product condition:
$$\overrightarrow{PQ}\!\cdot\!\overrightarrow{PR}=0 \;\Longrightarrow\; 6\cdot6+\frac{b^{2}}{a}\left(-\frac{b^{2}}{a}\right)=0 \;\Longrightarrow\; 36-\frac{b^{4}}{a^{2}}=0.$$ Hence
$$b^{4}=36a^{2}\;\Longrightarrow\; b^{2}=6a \;.(2)$$

Substitute $$b^{2}=6a$$ from $$(2)$$ into $$(1)$$:
$$a^{2}+6a-9=0.$$ Solving the quadratic,
$$a=\frac{-6\pm\sqrt{36+36}}{2}=\frac{-6\pm6\sqrt{2}}{2} =-3\pm3\sqrt{2}.$$ Since $$a\gt0,$$ take
$$a = -3+3\sqrt{2}=3(\sqrt{2}-1).$$

Now
$$b^{2}=6a=6\!\left[3(\sqrt{2}-1)\right]=18(\sqrt{2}-1).$$

The product required is
$$a^{2}b^{2} =\Bigl[\,3(\sqrt{2}-1)\Bigr]^{2}\,\,\Bigl[\,18(\sqrt{2}-1)\Bigr]$$ $$=9(3-2\sqrt{2})\times18(\sqrt{2}-1)$$ $$=162\,(3-2\sqrt{2})(\sqrt{2}-1).$$

Compute the remaining product:
$$(3-2\sqrt{2})(\sqrt{2}-1)=5\sqrt{2}-7.$$ Thus
$$a^{2}b^{2}=162\,(5\sqrt{2}-7)=810\sqrt{2}-1134.$$ Hence $$\alpha = 810,\quad \beta = 1134,\quad a^{2}b^{2}= \alpha\sqrt{2}-\beta.$$

The required sum is $$\alpha+\beta = 810+1134 = 1944.$$

Therefore, the value of $$\alpha+\beta$$ is $$1944$$.