Let $$y = f(x)$$ be a thrice differentiable function in $$(-5, 5)$$. Let the tangents to the curve $$y = f(x)$$ at $$(1, f(1))$$ and $$(3, f(3))$$ make angles $$\frac{\pi}{6}$$ and $$\frac{\pi}{4}$$, respectively with positive x-axis. If $$27\int_1^3 \{f'(t)\}^2 + 1\} f''(t) \, dt = \alpha + \beta\sqrt{3}$$ where $$\alpha, \beta$$ are integers, then the value of $$\alpha + \beta$$ equals

$$f'(1) = \tan(\pi/6) = 1/\sqrt{3}$$, $$f'(3) = \tan(\pi/4) = 1$$.

The integral: $$27\int_1^3 \{(f'(t))^2 + 1\}f''(t)dt$$.

Let $$u = f'(t)$$, $$du = f''(t)dt$$.

$$= 27\int_{f'(1)}^{f'(3)} (u^2+1)du = 27\left[\frac{u^3}{3}+u\right]_{1/\sqrt{3}}^{1}$$

At $$u = 1$$: $$\frac{1}{3} + 1 = \frac{4}{3}$$.

At $$u = 1/\sqrt{3}$$: $$\frac{1}{3 \cdot 3\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{1}{9\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{1+9}{9\sqrt{3}} = \frac{10}{9\sqrt{3}}$$.

$$= 27\left[\frac{4}{3} - \frac{10}{9\sqrt{3}}\right] = 27 \times \frac{4}{3} - 27 \times \frac{10}{9\sqrt{3}} = 36 - \frac{30}{\sqrt{3}} = 36 - 10\sqrt{3}$$

So $$\alpha = 36, \beta = -10$$. $$\alpha + \beta = 26$$.

The answer is Option (2): $$\boxed{26}$$.