Let the normals at all the points on a given curve pass through a fixed point $$(a, b)$$. If the curve passes through $$(3, -3)$$ and $$(4, -2\sqrt{2})$$, given that $$a - 2\sqrt{2}b = 3$$, then $$(a^2 + b^2 + ab)$$ is equal to ______.

If all normals to a curve pass through a fixed point $$(a, b)$$, then the curve is a circle with center $$(a, b)$$. Let the equation be $$(x - a)^2 + (y - b)^2 = r^2$$.

Since the curve passes through $$(3, -3)$$ and $$(4, -2\sqrt{2})$$, we have:

$$(3 - a)^2 + (-3 - b)^2 = (4 - a)^2 + (-2\sqrt{2} - b)^2$$

Expanding: $$9 - 6a + a^2 + 9 + 6b + b^2 = 16 - 8a + a^2 + 8 + 4\sqrt{2}b + b^2$$.

Simplifying: $$18 - 6a + 6b = 24 - 8a + 4\sqrt{2}b$$, which gives $$2a + (6 - 4\sqrt{2})b = 6$$.

We are also given $$a - 2\sqrt{2}b = 3$$, so $$a = 3 + 2\sqrt{2}b$$. Substituting into the first equation:

$$2(3 + 2\sqrt{2}b) + (6 - 4\sqrt{2})b = 6$$, which gives $$6 + 4\sqrt{2}b + 6b - 4\sqrt{2}b = 6$$, so $$6b = 0$$, hence $$b = 0$$.

Then $$a = 3 + 0 = 3$$, and $$a^2 + b^2 + ab = 9 + 0 + 0 = 9$$.