If $$\lim_{x \to 0} \frac{3 + \alpha \sin x + \beta \cos x + \log_e(1 - x)}{3\tan^2 x} = \frac{1}{3}$$, then $$2\alpha - \beta$$ is equal to :

$$\lim_{x \to 0} \frac{3 + \alpha\sin x + \beta\cos x + \ln(1-x)}{3\tan^2 x} = \frac{1}{3}$$

For the limit to be finite with $$3\tan^2 x \to 0$$, the numerator must also → 0:

At $$x = 0$$: $$3 + 0 + \beta + 0 = 0 \Rightarrow \beta = -3$$.

Now expand to second order:

$$\alpha\sin x = \alpha x - \frac{\alpha x^3}{6} + \ldots$$

$$\beta\cos x = -3(1 - \frac{x^2}{2} + \ldots) = -3 + \frac{3x^2}{2} + \ldots$$

$$\ln(1-x) = -x - \frac{x^2}{2} + \ldots$$

Numerator: $$3 + \alpha x - 3 + \frac{3x^2}{2} - x - \frac{x^2}{2} + O(x^3)$$

$$= (\alpha - 1)x + x^2 + O(x^3)$$

For the limit to be finite (denominator is $$3x^2$$), we need the $$x$$ coefficient to be 0:

$$\alpha - 1 = 0 \Rightarrow \alpha = 1$$.

Then: limit = $$\frac{x^2}{3x^2} = \frac{1}{3}$$. ✓

$$2\alpha - \beta = 2(1) - (-3) = 5$$.

The answer corresponds to Option (3).