Let $$M = \begin{bmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{bmatrix}$$ where $$a, b$$ and $$c$$ are real numbers such that $$a + b + c = 0$$ and $$abc \neq 0$$. If $$\det M = 0$$, then the maximum possible value of $$\frac{a^2 + b^2}{c^2}$$ is ____

Matrix $$M = \begin{bmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{bmatrix}$$ is a Vandermonde matrix, so $$\det M = (b-a)(c-a)(c-b)$$.

Since $$abc \neq 0$$, $$\det M = 0$$ forces two of $$a, b, c$$ to be equal. Combined with $$a + b + c = 0$$, there are three cases:

Case 1: $$a = b$$. Then $$c = -2a$$. So $$a^2 + b^2 = 2a^2$$ and $$c^2 = 4a^2$$. Ratio = $$\dfrac{2a^2}{4a^2} = \dfrac{1}{2}$$.

Case 2: $$b = c$$. Then $$a = -2c$$. So $$a^2 + b^2 = (-2c)^2 + c^2 = 4c^2 + c^2 = 5c^2$$. Ratio = $$\dfrac{5c^2}{c^2} = \mathbf{5}$$.

Case 3: $$a = c$$. Then $$b = -2c$$. So $$a^2 + b^2 = c^2 + 4c^2 = 5c^2$$. Ratio = $$\dfrac{5c^2}{c^2} = \mathbf{5}$$.

Maximum ratio = $$\mathbf{5}$$ (achieved in cases 2 and 3, where one of $$a, b$$ equals $$c$$, while the other equals $$-2c$$).