Let $$\alpha, \beta, \gamma$$ be the real roots of the equation, $$x^3 + ax^2 + bx + c = 0$$, $$(a, b, c \in R$$ and $$a, b \neq 0)$$. If the system of equations (in $$u, v, w$$) given by $$\alpha u + \beta v + \gamma w = 0$$, $$\beta u + \gamma v + \alpha w = 0$$, $$\gamma u + \alpha v + \beta w = 0$$ has non-trivial solution, then the value of $$\frac{a^2}{b}$$ is:

The system of equations in $$u, v, w$$ has a non-trivial solution, so the coefficient determinant must be zero: $$\begin{vmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix} = 0$$.

This is a circulant determinant. Expanding it: $$\alpha(\gamma\beta - \alpha^2) - \beta(\beta^2 - \alpha\gamma) + \gamma(\beta\alpha - \gamma^2)$$. Simplifying: $$\alpha\beta\gamma - \alpha^3 - \beta^3 + \alpha\beta\gamma + \alpha\beta\gamma - \gamma^3 = 3\alpha\beta\gamma - (\alpha^3 + \beta^3 + \gamma^3)$$.

Setting this to zero: $$\alpha^3 + \beta^3 + \gamma^3 = 3\alpha\beta\gamma$$. By the well-known identity, $$\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma = (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha)$$.

So either $$\alpha + \beta + \gamma = 0$$ or $$\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha = 0$$. The second condition means $$\frac{1}{2}[(\alpha-\beta)^2 + (\beta-\gamma)^2 + (\gamma-\alpha)^2] = 0$$, which requires $$\alpha = \beta = \gamma$$.

By Vieta's formulas for $$x^3 + ax^2 + bx + c = 0$$: $$\alpha + \beta + \gamma = -a$$, $$\alpha\beta + \beta\gamma + \gamma\alpha = b$$, $$\alpha\beta\gamma = -c$$.

Case 1: If $$\alpha + \beta + \gamma = 0$$, then $$a = 0$$, but we are told $$a \neq 0$$.

Case 2: If $$\alpha = \beta = \gamma$$, then $$\alpha + \beta + \gamma = 3\alpha = -a$$ and $$\alpha\beta + \beta\gamma + \gamma\alpha = 3\alpha^2 = b$$. So $$\frac{a^2}{b} = \frac{(3\alpha)^2}{3\alpha^2} = \frac{9\alpha^2}{3\alpha^2} = 3$$.

The answer is Option B: 3.