Mention the centre of mass of three particles which are not in line but have equal masses.

(N/A) Let the three particles have equal masses $m_1 = m_2 = m_3 = m$.
Let their position vectors be $\vec{r}_1$,$\vec{r}_2$,and $\vec{r}_3$.
The centre of mass $\vec{R}_{cm}$ is given by the formula:
$\vec{R}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3}{m_1 + m_2 + m_3}$
Since $m_1 = m_2 = m_3 = m$,we can substitute this into the equation:
$\vec{R}_{cm} = \frac{m(\vec{r}_1 + \vec{r}_2 + \vec{r}_3)}{3m} = \frac{\vec{r}_1 + \vec{r}_2 + \vec{r}_3}{3}$
This result shows that the centre of mass of three particles of equal mass is the centroid of the triangle formed by the positions of the three particles.