Three forces $$\vec{P}, \vec{Q}$$ and $$\vec{R}$$ acting along $$IA, IB$$ and $$IC$$, where $$I$$ is the incentre of a $$\triangle ABC$$, are in equilibrium. Then $$\vec{P} : \vec{Q} : \vec{R}$$ is

Since the three forces are in equilibrium and act along the concurrent lines $$IA$$, $$IB$$, and $$IC$$, Lami's theorem can be applied.

Therefore:

$$\frac{P}{\sin\angle BIC} = \frac{Q}{\sin\angle CIA} = \frac{R}{\sin\angle AIB}$$

For the incentre $$I$$ of a triangle:

$$\angle BIC = 180^\circ - \left(\frac{B}{2} + \frac{C}{2}\right) = 180^\circ - \left(90^\circ - \frac{A}{2}\right) = 90^\circ + \frac{A}{2}$$

$$\angle CIA = 90^\circ + \frac{B}{2}$$

$$\angle AIB = 90^\circ + \frac{C}{2}$$

Substituting these values into Lami's theorem:

$$P : Q : R = \sin\left(90^\circ + \frac{A}{2}\right) : \sin\left(90^\circ + \frac{B}{2}\right) : \sin\left(90^\circ + \frac{C}{2}\right)$$

Using the trigonometric identity $$\sin(90^\circ + \theta) = \cos\theta$$:

$$P : Q : R = \cos\frac{A}{2} : \cos\frac{B}{2} : \cos\frac{C}{2}$$

Hence, the correct option is A.