Given, Initial position vector, $\vec{r}_1 = 3\hat{i} + 2\hat{j} - 6\hat{k}$ Final position vector, $\vec{r}_2 = 14\hat{i} + 13\hat{j} + 9\hat{k}$ Force acting on the particle, $\vec{F} = 4\hat{i} + \hat{j} + 3\hat{k} \text{ N}$

The displacement vector $\vec{d}$ is given by: $\vec{d} = \vec{r}_2 - \vec{r}_1$ $\vec{d} = (14\hat{i} + 13\hat{j} + 9\hat{k}) - (3\hat{i} + 2\hat{j} - 6\hat{k})$ $\vec{d} = (14 - 3)\hat{i} + (13 - 2)\hat{j} + (9 + 6)\hat{k} = 11\hat{i} + 11\hat{j} + 15\hat{k}$

The work done $W$ by a constant force is the dot product of the force vector and the displacement vector : $W = \vec{F} \cdot \vec{d}$ $W = (4\hat{i} + \hat{j} + 3\hat{k}) \cdot (11\hat{i} + 11\hat{j} + 15\hat{k})$ $W = (4 \times 11) + (1 \times 11) + (3 \times 15)$ $W = 44 + 11 + 45 = 100 \text{ J}$

Thus, the work done is $100 \text{ J}$.