Three masses of `1kg`, `2kg`, and `3kg`, are placed at the vertices of an equilateral traingle of side `1m`. Find the gravitational potential energy of this system.
Take `G= 6.67xx10^(-11) N-m^(2)//kg^(2)`.

Explanation

To find the gravitational potential energy of the system consisting of three masses placed at the vertices of an equilateral triangle, we can follow these steps:

### Step 1: Identify the masses and their positions
We have three masses:
- Mass \( m_A = 1 \, \text{kg} \)
- Mass \( m_B = 2 \, \text{kg} \)
- Mass \( m_C = 3 \, \text{kg} \)

These masses are placed at the vertices of an equilateral triangle with a side length of \( 1 \, \text{m} \).

### Step 2: Write the formula for gravitational potential energy
The gravitational potential energy \( U \) between two point masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by:
\[ U = -\frac{G m_1 m_2}{r} \]
where \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).

### Step 3: Calculate the potential energy between each pair of masses
1. **Between \( m_B \) and \( m_C \)**:
\[ U_{BC} = -\frac{G m_B m_C}{1} = -\frac{6.67 \times 10^{-11} \times 2 \times 3}{1} = -6.67 \times 10^{-11} \times 6 = -4.002 \times 10^{-10} \, \text{J} \]

2. **Between \( m_A \) and \( m_B \)**:
\[ U_{AB} = -\frac{G m_A m_B}{1} = -\frac{6.67 \times 10^{-11} \times 1 \times 2}{1} = -6.67 \times 10^{-11} \times 2 = -1.334 \times 10^{-10} \, \text{J} \]

3. **Between \( m_A \) and \( m_C \)**:
\[ U_{AC} = -\frac{G m_A m_C}{1} = -\frac{6.67 \times 10^{-11} \times 1 \times 3}{1} = -6.67 \times 10^{-11} \times 3 = -2.001 \times 10^{-10} \, \text{J} \]

### Step 4: Calculate the total gravitational potential energy
Now, we sum the potential energies calculated for each pair:
\[ U_{\text{total}} = U_{AB} + U_{BC} + U_{AC} \]
Substituting the values:
\[ U_{\text{total}} = (-1.334 \times 10^{-10}) + (-4.002 \times 10^{-10}) + (-2.001 \times 10^{-10}) \]
\[ U_{\text{total}} = -7.337 \times 10^{-10} \, \text{J} \]

### Final Answer
The gravitational potential energy of the system is:
\[ U_{\text{total}} = -7.337 \times 10^{-10} \, \text{J} \]