The centre of mass of a body

Write the expression of centre of mass of a system of $'n'$ particles and derive the formula of force acting on its centre of mass. 

Three point particles of masses $1.0\; \mathrm{kg} .1 .5 \;\mathrm{kg}$ and $2.5\; kg$ are placed at three comers of a right angle triangle of sides $4.0\; \mathrm{cm}, 3.0 \;\mathrm{cm}$ and $5.0\; \mathrm{cm}$ as shown in the figure. The center of mass of the system is at a point

What is centre of mass ?

From a uniform disc of radius $R$, an equilateral triangle of side $\sqrt 3 \,R$ is cut as shown. The new position of centre of mass is :

The position vector of three particles of masses $1\, kg, 2\, kg$ and $3\, kg$ are $\overrightarrow {{r_1}}  = (\widehat i + 4\widehat j + \widehat k)\,m,\overrightarrow {{r_2}}  = (\widehat i + \widehat j + \widehat k)\,m$ and $\overrightarrow {{r_3}}  = (2\widehat i - \widehat j - 2\widehat k)\,m$  respectively. The position  vector of their centre of mass is