Step-by-step Solution:

To determine the work done when the particle is displaced from point A to point B under the action of constant forces \( \vec{P} \) and \( \vec{Q} \), we can proceed with the following steps: 1. Given Forces and Position Vectors: - Forces: \[ \vec{P} = 2\hat{i} - 5\hat{j} + 6\hat{k} \] \[ \vec{Q} = -\hat{i} + 2\hat{j} - \hat{k} \] - Position vectors: \[ \vec{A} = 4\hat{i} - 3\hat{j} - 2\hat{k} \] \[ \vec{B} = 6\hat{i} + \hat{j} - 3\hat{k} \] 2. Calculate the Resultant Force: The resultant force \( \vec{F} \) is the sum of \( \vec{P} \) and \( \vec{Q} \): \[ \vec{F} = \vec{P} + \vec{Q} = (2\hat{i} - 5\hat{j} + 6\hat{k}) + (-\hat{i} + 2\hat{j} - \hat{k}) \] \[ \vec{F} = (2 - 1)\hat{i} + (-5 + 2)\hat{j} + (6 - 1)\hat{k} \] \[ \vec{F} = \hat{i} - 3\hat{j} + 5\hat{k} \] 3. Calculate the Displacement Vector: The displacement vector \( \vec{d} \) from A to B is: \[ \vec{d} = \vec{B} - \vec{A} = (6\hat{i} + \hat{j} - 3\hat{k}) - (4\hat{i} - 3\hat{j} - 2\hat{k}) \] \[ \vec{d} = (6 - 4)\hat{i} + (1 - (-3))\hat{j} + (-3 - (-2))\hat{k} \] \[ \vec{d} = 2\hat{i} + 4\hat{j} - \hat{k} \] 4. Calculate the Work Done: The work done \( W \) is the dot product of the resultant force \( \vec{F} \) and the displacement vector \( \vec{d} \): \[ W = \vec{F} \cdot \vec{d} = (\hat{i} - 3\hat{j} + 5\hat{k}) \cdot (2\hat{i} + 4\hat{j} - \hat{k}) \] \[ W = (1 \times 2) + (-3 \times 4) + (5 \times -1) \] \[ W = 2 - 12 - 5 \] \[ W = -15 \text{ units} \] Therefore, the work done when the particle is displaced from A to B is: \[ \boxed{-15 \text{ units}} \]