Three particles of masses `1kg`, `2kg` and `3kg` are placed at the corners A, B and C respectively of an equilateral triangle ABC of edge `1m`. Find the distance of their centre of mass from A.

Explanation

To find the distance of the center of mass from point A in the given problem, we will follow these steps:

### Step 1: Set the Coordinate System
We will place point A at the origin (0, 0). The coordinates of points B and C will be determined based on the geometry of the equilateral triangle.

- **Coordinates**:
- A (0, 0)
- B (1, 0) (since B is 1 meter to the right of A)
- C (0.5, \( \frac{\sqrt{3}}{2} \)) (the height of the triangle can be calculated using the formula for the height of an equilateral triangle)

### Step 2: Identify Masses
The masses at each point are:
- \( m_A = 1 \, \text{kg} \) at A
- \( m_B = 2 \, \text{kg} \) at B
- \( m_C = 3 \, \text{kg} \) at C

### Step 3: Calculate the Center of Mass Coordinates
The coordinates of the center of mass (CM) can be calculated using the formula:

\[ x_{cm} = \frac{m_A x_A + m_B x_B + m_C x_C}{m_A + m_B + m_C} \]

Substituting the values:

\[ x_{cm} = \frac{(1 \cdot 0) + (2 \cdot 1) + (3 \cdot 0.5)}{1 + 2 + 3} = \frac{0 + 2 + 1.5}{6} = \frac{3.5}{6} = \frac{7}{12} \, \text{m} \]

Now for the y-coordinate:

\[ y_{cm} = \frac{m_A y_A + m_B y_B + m_C y_C}{m_A + m_B + m_C} \]

Substituting the values:

\[ y_{cm} = \frac{(1 \cdot 0) + (2 \cdot 0) + (3 \cdot \frac{\sqrt{3}}{2})}{1 + 2 + 3} = \frac{0 + 0 + \frac{3\sqrt{3}}{2}}{6} = \frac{\frac{3\sqrt{3}}{2}}{6} = \frac{\sqrt{3}}{4} \, \text{m} \]

### Step 4: Calculate the Distance from A to the Center of Mass
The distance \( D \) from point A to the center of mass can be calculated using the Pythagorean theorem:

\[ D = \sqrt{x_{cm}^2 + y_{cm}^2} \]

Substituting the values we found:

\[ D = \sqrt{\left(\frac{7}{12}\right)^2 + \left(\frac{\sqrt{3}}{4}\right)^2} \]

Calculating each term:

\[ D = \sqrt{\frac{49}{144} + \frac{3}{16}} \]

To add these fractions, we need a common denominator. The least common multiple of 144 and 16 is 144:

\[ \frac{3}{16} = \frac{27}{144} \]

Thus, we have:

\[ D = \sqrt{\frac{49 + 27}{144}} = \sqrt{\frac{76}{144}} = \frac{\sqrt{76}}{12} \]

We can simplify \( \sqrt{76} \) as \( \sqrt{4 \cdot 19} = 2\sqrt{19} \):

\[ D = \frac{2\sqrt{19}}{12} = \frac{\sqrt{19}}{6} \, \text{m} \]

### Final Answer
The distance of the center of mass from point A is:

\[ \boxed{\frac{\sqrt{19}}{6} \, \text{m}} \]