three particles each of mass m are placed at three corner of an equilateral triangle of length l. find the position of center of mass in terms of coordinates

Taking origin at vertex B and taking axis along the side BC .
so the coordinates of the B are (0,0) and that of C are (a,0)
As the angle between the sides of an equilateral triangle is 60^o , so for point A the x-coordinate will be a cos60^o =$\frac{a}{2}$
x-coordinate will be a sin60^o= $\frac{\sqrt{3}a}{2}$

The length of the side of equilateral triangle is taken to be equal to a .
The coordinates of centre of mass are as follows:
$x_{cm}=\frac{m\times 0+m\times a+m\times \left({\displaystyle \frac{a}{2}}\right)}{3m}=\frac{a}{2}$
$y_{cm}=\frac{m\times 0+m\times 0+m\times \left({\displaystyle \frac{\sqrt{3}a}{2}}\right)}{3m}=\frac{a}{2\sqrt{3}}$