Three particles of masses 50g, 100g, and 150g are placed at the vertices of an equilateral triangle of side 1m. The (x, y) coordinates of the centre of mass will be:

Correct option is C (712m,34m)
The co-ordinates of the centre of mass
(\vec{r}{cm} = \frac{0 + 150 \times (\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j}) + 100 \times \hat{i}}{300})
(\vec{r}{cm} = \frac{7}{12}\hat{i} + \frac{3}{4}\hat{j})
(\therefore \text{Co-ordinate } (\frac{7}{12}, \frac{3}{4})m)