Centre of mass of three particles of masses `1 kg, 2 kg and 3 kg` lies at the point `(1,2,3)` and centre of mass of another system of particles `3 kg and 3 kg` lies at the point `(-1, 3, -2)`. Where should we put a particle of mass `5 kg` so that the centre of mass of entire system lies at the centre of mass of first system ?

Options

A. `(0, 0,0)`
B. `(1, 3,2)`
C. `(-1, 2,3)`
D. `(3, 1, 8)`

Explanation

To find the position where we should place a particle of mass 5 kg so that the center of mass of the entire system lies at the center of mass of the first system (1, 2, 3), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given data:**
- Masses of the first system: \( m_1 = 1 \, \text{kg}, m_2 = 2 \, \text{kg}, m_3 = 3 \, \text{kg} \)
- Center of mass of the first system: \( (x_{cm1}, y_{cm1}, z_{cm1}) = (1, 2, 3) \)
- Masses of the second system: \( m_4 = 3 \, \text{kg}, m_5 = 3 \, \text{kg} \)
- Center of mass of the second system: \( (x_{cm2}, y_{cm2}, z_{cm2}) = (-1, 3, -2) \)
- Mass of the additional particle: \( m_6 = 5 \, \text{kg} \)

2. **Calculate the total mass of the second system:**
\[ M_2 = m_4 + m_5 = 3 + 3 = 6 \, \text{kg} \]

3. **Set the center of mass of the second system:**
The center of mass of the second system is given by:
\[ x_{cm2} = \frac{m_4 \cdot x_4 + m_5 \cdot x_5}{M_2} \quad \text{and similar for } y \text{ and } z \]
Here, we can take the center of mass coordinates as \( (-1, 3, -2) \).

4. **Combine the two systems with the new mass:**
We need to find the coordinates \( (x, y, z) \) for the 5 kg mass such that the overall center of mass of the entire system (including the new mass) is \( (1, 2, 3) \).

5. **Total mass of the entire system:**
\[ M = m_1 + m_2 + m_3 + m_4 + m_5 + m_6 = 1 + 2 + 3 + 3 + 3 + 5 = 17 \, \text{kg} \]

6. **Set up the equations for center of mass:**
The center of mass equations for \( x \), \( y \), and \( z \) can be expressed as:
\[ x_{cm} = \frac{m_1 \cdot x_1 + m_2 \cdot x_2 + m_3 \cdot x_3 + m_4 \cdot x_4 + m_5 \cdot x_5 + m_6 \cdot x}{M} \]
\[ y_{cm} = \frac{m_1 \cdot y_1 + m_2 \cdot y_2 + m_3 \cdot y_3 + m_4 \cdot y_4 + m_5 \cdot y_5 + m_6 \cdot y}{M} \]
\[ z_{cm} = \frac{m_1 \cdot z_1 + m_2 \cdot z_2 + m_3 \cdot z_3 + m_4 \cdot z_4 + m_5 \cdot z_5 + m_6 \cdot z}{M} \]

7. **Substituting known values:**
For \( x \):
\[ 1 = \frac{1 \cdot 1 + 2 \cdot 2 + 3 \cdot 3 + 3 \cdot (-1) + 3 \cdot (-1) + 5 \cdot x}{17} \]
Simplifying gives:
\[ 1 = \frac{1 + 4 + 9 - 3 - 3 + 5x}{17} \]
\[ 1 = \frac{8 + 5x}{17} \]
\[ 17 = 8 + 5x \implies 5x = 9 \implies x = \frac{9}{5} = 1.8 \]

For \( y \):
\[ 2 = \frac{1 \cdot 2 + 2 \cdot 2 + 3 \cdot 3 + 3 \cdot 3 + 3 \cdot 3 + 5 \cdot y}{17} \]
Simplifying gives:
\[ 2 = \frac{2 + 4 + 9 + 9 + 9 + 5y}{17} \]
\[ 34 = 33 + 5y \implies 5y = 1 \implies y = \frac{1}{5} = 0.2 \]

For \( z \):
\[ 3 = \frac{1 \cdot 3 + 2 \cdot 3 + 3 \cdot 3 + 3 \cdot (-2) + 3 \cdot (-2) + 5 \cdot z}{17} \]
Simplifying gives:
\[ 3 = \frac{3 + 6 + 9 - 6 - 6 + 5z}{17} \]
\[ 51 = 6 + 5z \implies 5z = 45 \implies z = 9 \]

8. **Final coordinates for the 5 kg mass:**
The coordinates where the 5 kg mass should be placed are:
\[ (x, y, z) = (1.8, 0.2, 9) \]

### Final Answer:
The 5 kg mass should be placed at the coordinates \( (1.8, 0.2, 9) \).