Let $$\alpha$$ be a root of the equation $$(a-c)x^2 + (b-a)x + (c-b) = 0$$ where $$a, b, c$$ are distinct real numbers such that the matrix $$\begin{pmatrix} \alpha^2 & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c \end{pmatrix}$$ is singular. Then the value of $$\frac{(a-c)^2}{(b-a)(c-b)} + \frac{(b-a)^2}{(a-c)(c-b)} + \frac{(c-b)^2}{(a-c)(b-a)}$$ is

The equation $$(a-c)x^2 + (b-a)x + (c-b) = 0$$ has $$x = 1$$ as a root (since the sum of coefficients is zero), and by Vieta’s formulas the other root is $$\alpha = \frac{c-b}{a-c}$$.

The given matrix is singular for either root; in particular, when $$\alpha = 1$$ the first two rows coincide, so the determinant vanishes.

Setting $$p = a - c$$, $$q = b - a$$ and $$r = c - b$$, one observes that $$p + q + r = 0$$. The expression then becomes $$\frac{p^2}{qr} + \frac{q^2}{pr} + \frac{r^2}{pq} = \frac{p^3 + q^3 + r^3}{pqr}$$. Since $$p + q + r = 0$$, the identity $$p^3 + q^3 + r^3 = 3pqr$$ applies, giving $$\frac{3pqr}{pqr} = 3$$.

The correct answer is Option 2: 3.