Two particles of mass \(5~\text{kg}\) and \(10~\text{kg}\) respectively are attached to the two ends of a rigid rod of length \(1~\text{m}\) with negligible mass. The centre of mass of the system from the \(5~\text{kg}\) particle is nearly at a distance of:
1. \(50~\text{cm}\)
2. \(67~\text{cm}\)
3. \(80~\text{cm}\)
4. \(33~\text{cm}\)

Three identical spheres, each of mass \(M\), are placed at the corners of a right-angle triangle with mutually perpendicular sides equal to \(2~\text{m}\) (see figure). Taking the point of intersection of the two mutually perpendicular sides as the origin, find the position vector of the centre of mass. 
        
1. \(2(\hat{i}+\hat{j})\)
2. \(\hat{i}+\hat{j}\)
3. \(\frac{2}{3}(\hat{i}+\hat{j})\)
4. \(\frac{4}{3}(\hat{i}+\hat{j})\)