The product of all the rational roots of the equation $$(x^2 - 9x + 11)^2 - (x - 4)(x - 5) = 3$$ is equal to:

$$(x^2 - 9x + 11)^2 - (x - 4)(x - 5) = 3$$

$$(x-4)(x-5)=x^2-9x+20$$

Let $$x^2-9x+20=t$$

$$\left(t-9\right)^2-t=3$$

$$t^2-19t+78=0$$

$$\left(t-13\right)\left(t-6\right)=0$$

So, $$t=13\ ,\ 6$$

Now, two cases will be formed: 

Case 1: $$t=13$$

$$x^2-9x+20=13$$

$$x^2-9x+7=0$$

The discriminant is $$\sqrt{53}$$. So, all the roots will be irrational. 

Case 2: $$t=6$$

$$x^2-9x+20=6$$

$$x^2-9x+14=0$$

$$\left(x-7\right)\left(x-2\right)=0$$

$$x=2\ ,\ 7$$

These roots are rational. 

We have to find the product of rational roots. 

Product = $$\left(2\times7\right)=14$$

$$\therefore\ $$ The required answer is A.