If $$(a, b)$$ be the orthocentre of the triangle whose vertices are $$(1, 2), (2, 3)$$ and $$(3, 1)$$, and $$I_1 = \int_a^b x \sin(4x - x^2) \, dx$$, $$I_2 = \int_a^b \sin(4x - x^2) \, dx$$, then $$36 \frac{I_1}{I_2}$$ is equal to :

First, find the orthocentre of the triangle with vertices $$A(1,2)$$, $$B(2,3)$$, $$C(3,1)$$.

Slope of BC = $$\frac{1-3}{3-2} = -2$$. Altitude from A perpendicular to BC has slope $$\frac{1}{2}$$.

Altitude from A: $$y - 2 = \frac{1}{2}(x - 1)$$, i.e., $$2y - 4 = x - 1$$, i.e., $$x = 2y - 3$$.

Slope of AC = $$\frac{1-2}{3-1} = -\frac{1}{2}$$. Altitude from B perpendicular to AC has slope $$2$$.

Altitude from B: $$y - 3 = 2(x - 2)$$, i.e., $$y = 2x - 1$$.

Intersection: $$x = 2(2x-1) - 3 = 4x - 5$$, so $$-3x = -5$$, $$x = 5/3$$.

$$y = 2(5/3) - 1 = 7/3$$.

Orthocentre: $$(a, b) = (5/3, 7/3)$$. Note $$a + b = 4$$.

$$I_1 = \int_{5/3}^{7/3} x\sin(4x - x^2) dx$$, $$I_2 = \int_{5/3}^{7/3} \sin(4x - x^2) dx$$.

Let $$u = 4x - x^2$$. Note that $$4x - x^2 = -(x-2)^2 + 4$$, symmetric about $$x = 2$$.

The interval $$[5/3, 7/3]$$ is symmetric about $$x = 2$$.

Using the property: for $$f(a+b-x) = f(x)$$ (where $$a + b = 5/3 + 7/3 = 4$$):

$$4(a+b-x) - (a+b-x)^2 = 4(4-x) - (4-x)^2 = 16 - 4x - 16 + 8x - x^2 = 4x - x^2$$

So $$\sin(4(4-x)-(4-x)^2) = \sin(4x - x^2)$$.

By the substitution $$x \to 4 - x$$:

$$I_1 = \int_{5/3}^{7/3} (4-x)\sin(4x-x^2) dx = 4I_2 - I_1$$

$$2I_1 = 4I_2$$

$$\frac{I_1}{I_2} = 2$$

$$36 \cdot \frac{I_1}{I_2} = 36 \times 2 = 72$$

The answer is $$\boxed{72}$$, which corresponds to Option (1).