Let's analyze the situation step by step.

The particle’s position vector is given by

$$\vec{r} = a(\hat{i}\cos\omega t + \hat{j}\sin\omega t),$$

which represents uniform circular motion in the xy-plane.

Its velocity is the time derivative of the position:

$$\vec{v} = \frac{d\vec{r}}{dt} = -a\omega\sin\omega t\,\hat{i} + a\omega\cos\omega t\,\hat{j}.$$

The acceleration, obtained by differentiating the velocity, is:

$$\vec{a} = \frac{d\vec{v}}{dt} = -a\omega^2\cos\omega t\,\hat{i} - a\omega^2\sin\omega t\,\hat{j}.$$

Notice that this can be rewritten as:

$$\vec{a} = -\omega^2 \vec{r}.$$

This indicates that the acceleration (and hence the force, since $$\vec{F}=m\vec{a}$$) is directed opposite to the position vector $$\vec{r}$$—that is, radially inward.

To check against the given options:

Option A (Opposite to $$\vec{L}$$): The angular momentum $$\vec{L}=m\vec{r}\times\vec{v}$$ is perpendicular to the plane of motion (along $$\pm\hat{k}$$) and does not indicate a direction in the plane.

Option B (Opposite to $$\vec{L}\times\vec{P}$$): For a particle in circular motion, when you compute $$\vec{L}\times\vec{P}$$ (with $$\vec{P} = m\vec{v}$$), you’ll find that it is proportional to $$-\vec{r}.$$ Thus, the direction opposite to $$\vec{L}\times\vec{P}$$ would be the same as $$\vec{r}$$, which is the outward radial direction—not matching the inward (centripetal) force.

Option D (Opposite to $$\vec{P}$$): The linear momentum is tangential; the force (centripetal) is not tangential but directed radially inward.

Option C (Opposite to $$\vec{r}$$): As we found, the acceleration (and hence force) is given by $$\vec{F}= m\vec{a} = -m\omega^2 \vec{r},$$ which is clearly opposite to $$\vec{r}.$$

Therefore, the force is directed opposite to $$\vec{r},$$ making Option C the correct answer.