The position vector of a particle is given by r = 3 \,t^2 i + 2 \,t...

Given position vector $\vec{r} = \sqrt{3}t^2\hat{i} + \sqrt{2}t\hat{j} + \sqrt{5}\hat{k}$

Velocity $\vec{v} = \dfrac{d\vec{r}}{dt} = 2\sqrt{3}t\hat{i} + \sqrt{2}\hat{j}$

Instantaneous momentum $\vec{p} = m\vec{v} = m(2\sqrt{3}t\hat{i} + \sqrt{2}\hat{j})$

Acceleration $\vec{a} = \dfrac{d\vec{v}}{dt} = 2\sqrt{3}\hat{i}$

Force $\vec{F} = m\vec{a} = 2\sqrt{3}m\hat{i}$

Taking the cross product of force and momentum:

$\vec{F} \times \vec{p} = (2\sqrt{3}m\hat{i}) \times m(2\sqrt{3}t\hat{i} + \sqrt{2}\hat{j}) = 2\sqrt{6}m^2\hat{k} \neq 0$

Since the cross product is non-zero, the force is not parallel to the instantaneous momentum.

Also, $\vec{F} \cdot \vec{p} = 12m^2t \neq 0$, so they are not perpendicular either.

Answer: The force acting on the particle is not parallel to the instantaneous momentum of the particle.