Find the velocity of the centre of mass of a system of particles of masses 1 kg and 3 kg which are at `(2hati + 5hatj + 13hatk)`m and `(-6hati + 4hatj - 2hatk)` possessing velocities `(10hati - 7hatj - 3hatk)ms^(-1)` and `(7hati - 9hatj + 6hatk)ms^(-1)` respectively.

Answer

To find the velocity of the center of mass of a system of particles, we can follow these steps:

### Step 1: Identify the masses and velocities
We have two particles:
- Mass \( m_1 = 1 \, \text{kg} \) at position \( \vec{r_1} = 2\hat{i} + 5\hat{j} + 13\hat{k} \, \text{m} \) with velocity \( \vec{v_1} = 10\hat{i} - 7\hat{j} - 3\hat{k} \, \text{m/s} \).
- Mass \( m_2 = 3 \, \text{kg} \) at position \( \vec{r_2} = -6\hat{i} + 4\hat{j} - 2\hat{k} \, \text{m} \) with velocity \( \vec{v_2} = 7\hat{i} - 9\hat{j} + 6\hat{k} \, \text{m/s} \).

### Step 2: Calculate the total mass
The total mass \( M \) of the system is given by:
\[ M = m_1 + m_2 = 1 \, \text{kg} + 3 \, \text{kg} = 4 \, \text{kg} \]

### Step 3: Calculate the momentum of each mass
The momentum \( \vec{p} \) of each mass is given by:
\[ \vec{p_1} = m_1 \vec{v_1} = 1 \cdot (10\hat{i} - 7\hat{j} - 3\hat{k}) = 10\hat{i} - 7\hat{j} - 3\hat{k} \]
\[ \vec{p_2} = m_2 \vec{v_2} = 3 \cdot (7\hat{i} - 9\hat{j} + 6\hat{k}) = 21\hat{i} - 27\hat{j} + 18\hat{k} \]

### Step 4: Calculate the total momentum of the system
The total momentum \( \vec{P} \) of the system is:
\[ \vec{P} = \vec{p_1} + \vec{p_2} = (10\hat{i} - 7\hat{j} - 3\hat{k}) + (21\hat{i} - 27\hat{j} + 18\hat{k}) \]
\[ \vec{P} = (10 + 21)\hat{i} + (-7 - 27)\hat{j} + (-3 + 18)\hat{k} = 31\hat{i} - 34\hat{j} + 15\hat{k} \]

### Step 5: Calculate the velocity of the center of mass
The velocity of the center of mass \( \vec{v_{cm}} \) is given by:
\[ \vec{v_{cm}} = \frac{\vec{P}}{M} = \frac{31\hat{i} - 34\hat{j} + 15\hat{k}}{4} \]
\[ \vec{v_{cm}} = \frac{31}{4}\hat{i} - \frac{34}{4}\hat{j} + \frac{15}{4}\hat{k} \]

### Final Answer
Thus, the velocity of the center of mass is:
\[ \vec{v_{cm}} = 7.75\hat{i} - 8.5\hat{j} + 3.75\hat{k} \, \text{m/s} \]