If $$\lim_{x \to 0} \frac{\alpha e^x + \beta e^{-x} + \gamma \sin x}{x \sin^2 x} = \frac{2}{3}$$, where $$\alpha, \beta, \gamma \in \mathbb{R}$$, then which of the following is NOT correct?

We need to find $$\alpha, \beta, \gamma$$ such that $$\displaystyle\lim_{x \to 0} \frac{\alpha e^x + \beta e^{-x} + \gamma \sin x}{x \sin^2 x} = \frac{2}{3}$$.

We use Taylor expansions about $$x = 0$$. We have $$e^x = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \cdots$$, and $$e^{-x} = 1 - x + \dfrac{x^2}{2} - \dfrac{x^3}{6} + \cdots$$, and $$\sin x = x - \dfrac{x^3}{6} + \cdots$$.

The numerator becomes $$\alpha\!\left(1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \cdots\right) + \beta\!\left(1 - x + \dfrac{x^2}{2} - \dfrac{x^3}{6} + \cdots\right) + \gamma\!\left(x - \dfrac{x^3}{6} + \cdots\right)$$.

Collecting by powers of $$x$$: the constant term is $$\alpha + \beta$$, the coefficient of $$x$$ is $$\alpha - \beta + \gamma$$, the coefficient of $$x^2$$ is $$\dfrac{\alpha + \beta}{2}$$, and the coefficient of $$x^3$$ is $$\dfrac{\alpha - \beta}{6} - \dfrac{\gamma}{6} = \dfrac{\alpha - \beta - \gamma}{6}$$.

The denominator is $$x \sin^2 x \approx x \cdot x^2 = x^3$$ as $$x \to 0$$. For the limit to be finite and non-zero, the numerator must begin at $$x^3$$, so all lower-order coefficients must vanish.

Setting the constant term to zero: $$\alpha + \beta = 0$$ ... (i). This also makes the $$x^2$$ coefficient vanish automatically. Setting the coefficient of $$x$$ to zero: $$\alpha - \beta + \gamma = 0$$ ... (ii).

From (i), $$\beta = -\alpha$$. Substituting into (ii): $$\alpha + \alpha + \gamma = 0$$, so $$\gamma = -2\alpha$$.

Now the leading term of the numerator is $$\dfrac{\alpha - \beta - \gamma}{6} \cdot x^3 = \dfrac{\alpha + \alpha + 2\alpha}{6} \cdot x^3 = \dfrac{4\alpha}{6} \cdot x^3 = \dfrac{2\alpha}{3} \cdot x^3$$.

Hence the limit equals $$\dfrac{2\alpha/3}{1} = \dfrac{2\alpha}{3}$$. Setting this equal to $$\dfrac{2}{3}$$ gives $$\alpha = 1$$. Therefore $$\beta = -1$$ and $$\gamma = -2$$.

Now we check each option:

Option A: $$\alpha^2 + \beta^2 + \gamma^2 = 1 + 1 + 4 = 6$$ ✓ (correct statement)

Option B: $$\alpha\beta + \beta\gamma + \gamma\alpha + 1 = (1)(-1) + (-1)(-2) + (-2)(1) + 1 = -1 + 2 - 2 + 1 = 0$$ ✓ (correct statement)

Option C: $$\alpha\beta^2 + \beta\gamma^2 + \gamma\alpha^2 + 3 = (1)(1) + (-1)(4) + (-2)(1) + 3 = 1 - 4 - 2 + 3 = -2 \neq 0$$ ✗ (NOT correct)

Option D: $$\alpha^2 - \beta^2 + \gamma^2 = 1 - 1 + 4 = 4$$ ✓ (correct statement)

Hence, the correct answer is Option C.