Two bodies of masses 10 kg and 2 kg are moving with velocities `2 hat(i)-7 hat(j)+3 hat(k)` and `-10 hat(i)+35 hat(j)-3hat(j) ms^(-1)` respectively. Find the velocity of the centre of mass of the system.

Answer

To find the velocity of the center of mass (V_cm) of the system consisting of two bodies with given masses and velocities, we can use the formula:

\[ V_{cm} = \frac{m_1 \vec{v_1} + m_2 \vec{v_2}}{m_1 + m_2} \]

Where:
- \( m_1 \) and \( m_2 \) are the masses of the two bodies,
- \( \vec{v_1} \) and \( \vec{v_2} \) are the velocities of the two bodies.

### Step 1: Identify the masses and velocities
- Mass of the first body, \( m_1 = 10 \, \text{kg} \)
- Velocity of the first body, \( \vec{v_1} = 2 \hat{i} - 7 \hat{j} + 3 \hat{k} \, \text{ms}^{-1} \)
- Mass of the second body, \( m_2 = 2 \, \text{kg} \)
- Velocity of the second body, \( \vec{v_2} = -10 \hat{i} + 35 \hat{j} - 3 \hat{k} \, \text{ms}^{-1} \)

### Step 2: Calculate the total mass
\[ m_{total} = m_1 + m_2 = 10 \, \text{kg} + 2 \, \text{kg} = 12 \, \text{kg} \]

### Step 3: Calculate the momentum of each body
- Momentum of the first body:
\[ m_1 \vec{v_1} = 10 \, \text{kg} \cdot (2 \hat{i} - 7 \hat{j} + 3 \hat{k}) = (20 \hat{i} - 70 \hat{j} + 30 \hat{k}) \, \text{kg m/s} \]

- Momentum of the second body:
\[ m_2 \vec{v_2} = 2 \, \text{kg} \cdot (-10 \hat{i} + 35 \hat{j} - 3 \hat{k}) = (-20 \hat{i} + 70 \hat{j} - 6 \hat{k}) \, \text{kg m/s} \]

### Step 4: Sum the momenta
\[ \text{Total momentum} = m_1 \vec{v_1} + m_2 \vec{v_2} \]
\[ = (20 \hat{i} - 70 \hat{j} + 30 \hat{k}) + (-20 \hat{i} + 70 \hat{j} - 6 \hat{k}) \]
\[ = (20 - 20) \hat{i} + (-70 + 70) \hat{j} + (30 - 6) \hat{k} \]
\[ = 0 \hat{i} + 0 \hat{j} + 24 \hat{k} = 24 \hat{k} \, \text{kg m/s} \]

### Step 5: Calculate the velocity of the center of mass
\[ V_{cm} = \frac{\text{Total momentum}}{m_{total}} = \frac{24 \hat{k}}{12} = 2 \hat{k} \, \text{ms}^{-1} \]

### Final Answer
The velocity of the center of mass of the system is:
\[ \boxed{2 \hat{k} \, \text{ms}^{-1}} \]