Find the center of mass of three particles of masses m, 2m and 3m which are at the vertices of an equilateral triangle of sides a.

Choosing the axes as shown in figure and applying the formulae of centre of

Mass X = ∑m_ix_i/M ,  Y = ∑m_iy_i/M ,  Z = ∑m_iz_i/M

Therefore we have,

x_cm = (0 + 2ma + 3ma cos 60°)/(m + 2m + 3m)= 7/12 a ;  y_cm = (0 + 0 + 3ma sin 60°)/6m=1/4√3a

Z_cm = 0