The position vector of the centre of mass $\vec{R}_{CM}$ for a system of two particles is given by: $\vec{R}_{CM} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$ Given $m_1 = 1 \text{ kg}$, $\vec{r}_1 = \hat{i}+2\hat{j}+\hat{k}$ and $m_2 = 3 \text{ kg}$, $\vec{r}_2 = -3\hat{i}-2\hat{j}+\hat{k}$. Substituting the given values: $\vec{R}_{CM} = \frac{1(\hat{i}+2\hat{j}+\hat{k}) + 3(-3\hat{i}-2\hat{j}+\hat{k})}{1 + 3}$ $\vec{R}_{CM} = \frac{\hat{i}+2\hat{j}+\hat{k} - 9\hat{i}-6\hat{j}+3\hat{k}}{4}$ $\vec{R}_{CM} = \frac{-8\hat{i}-4\hat{j}+4\hat{k}}{4} = -2\hat{i} - \hat{j} + \hat{k}$