Let $$a, b, c$$ be in arithmetic progression. Let the centroid of the triangle with vertices $$(a, c)$$, $$(2, b)$$ and $$(a, b)$$ be $$\left(\frac{10}{3}, \frac{7}{3}\right)$$. If $$\alpha, \beta$$ are the roots of the equation $$ax^2 + bx + 1 = 0$$, then the value of $$\alpha^2 + \beta^2 - \alpha\beta$$ is:

We are given that $$a, b, c$$ are in arithmetic progression, so $$2b = a + c$$. The centroid of the triangle with vertices $$(a, c)$$, $$(2, b)$$, and $$(a, b)$$ is $$\left(\frac{10}{3}, \frac{7}{3}\right)$$.

From the centroid formula, the x-coordinate gives us $$\frac{a + 2 + a}{3} = \frac{10}{3}$$, which simplifies to $$2a + 2 = 10$$, so $$a = 4$$.

The y-coordinate gives us $$\frac{c + b + b}{3} = \frac{7}{3}$$, which simplifies to $$c + 2b = 7$$.

Since $$a, b, c$$ are in AP, we have $$2b = a + c = 4 + c$$, so $$c = 2b - 4$$. Substituting into $$c + 2b = 7$$, we get $$(2b - 4) + 2b = 7$$, giving $$4b = 11$$, so $$b = \frac{11}{4}$$ and $$c = \frac{11}{2} - 4 = \frac{3}{2}$$.

Now the equation $$ax^2 + bx + 1 = 0$$ becomes $$4x^2 + \frac{11}{4}x + 1 = 0$$. By Vieta's formulas, $$\alpha + \beta = -\frac{b}{a} = -\frac{11/4}{4} = -\frac{11}{16}$$ and $$\alpha\beta = \frac{1}{a} = \frac{1}{4}$$.

We need $$\alpha^2 + \beta^2 - \alpha\beta = (\alpha + \beta)^2 - 2\alpha\beta - \alpha\beta = (\alpha + \beta)^2 - 3\alpha\beta$$.

Substituting the values, we get $$\left(-\frac{11}{16}\right)^2 - 3 \cdot \frac{1}{4} = \frac{121}{256} - \frac{192}{256} = -\frac{71}{256}$$.

Therefore, the value of $$\alpha^2 + \beta^2 - \alpha\beta$$ is $$-\frac{71}{256}$$.