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(D) The formula for the position vector of the centre of mass is given by $\overrightarrow{r}_{cm} = \frac{m_{1} \overrightarrow{r}_{1} + m_{2} \overr

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$-\hat{i} + 3\hat{j} - 2\hat{k}$

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(D) The formula for the position vector of the centre of mass is given by $\overrightarrow{r}_{cm} = \frac{m_{1} \overrightarrow{r}_{1} + m_{2} \overrightarrow{r}_{2}}{m_{1} + m_{2}}$.Given $m_{1} = 1\,kg$,$\overrightarrow{r}_{1} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $m_{2} = 3\,kg$,$\overrightarrow{r}_{2} = -2\hat{i} + 3\hat{j} - 4\hat{k}$.Substituting the values:$\overrightarrow{r}_{cm} = \frac{1(2\hat{i} + 3\hat{j} + 4\hat{k}) + 3(-2\hat{i} + 3\hat{j} - 4\hat{k})}{1 + 3}$$\overrightarrow{r}_{cm} = \frac{2\hat{i} + 3\hat{j} + 4\hat{k} - 6\hat{i} + 9\hat{j} - 12\hat{k}}{4}$$\overrightarrow{r}_{cm} = \frac{-4\hat{i} + 12\hat{j} - 8\hat{k}}{4}$$\overrightarrow{r}_{cm} = -\hat{i} + 3\hat{j} - 2\hat{k}$.

Two particles of masses $1\,kg$ and $3\,kg$ have position vectors $2\hat{i} + 3\hat{j} + 4\hat{k}$ and $-2\hat{i} + 3\hat{j} - 4\hat{k}$ respectively. The position vector of the centre of mass is:

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