Particles of masses 100 and 300 gram have position vectors $\left(2\hat{\mathrm{i}}+5\hat{\mathrm{j}}+13\hat{\mathrm{k}}\right)$ and $\left(-6\hat{\mathrm{i}}+4\hat{\mathrm{j}}+2\hat{\mathrm{k}}\right)$ . Position vector of their centre of mass is

Let the two position vectors $r_1 and r_2$  as we have two of them.

Given,

${\mathrm{r}}_1=2\hat{\mathrm{i}}+5\hat{\mathrm{j}}+13\hat{\mathrm{k}}, {\mathrm{r}}_2=-6\hat{\mathrm{i}}+4\hat{\mathrm{j}}-2\hat{\mathrm{k}}$

First, we shall determine where the centre of mass is right now. The equation for the centre of mass's instantaneous position is as follows:

${\mathrm{r}}_{\mathrm{com}}=\frac{{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{m}}_{\mathrm{i}}{\mathrm{r}}_{\mathrm{i}}}}{{\displaystyle \underset{\mathrm{i}=1}{\overset{\mathrm{n}}{\sum {\mathrm{m}}_{\mathrm{i}}}}}}=\frac{{\mathrm{m}}_1{\mathrm{r}}_1+{\mathrm{m}}_2{\mathrm{r}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}$

${\mathrm{r}}_{\mathrm{com}}=\frac{100(2\hat{\mathrm{i}}+5\hat{\mathrm{j}}+13\hat{\mathrm{k}})+300(-6\hat{\mathrm{i}}+4\hat{\mathrm{j}}+-2\hat{\mathrm{k}})}{400}$

${\mathrm{r}}_{\mathrm{com}}=\frac{-16}{4}\hat{\mathrm{i}}+\frac{17}{4}\hat{\mathrm{j}}+\frac{7}{4}\hat{\mathrm{k}}$

Hence the correct answer is $\frac{-16}{4}\hat{\mathrm{i}}+\frac{17}{4}\hat{\mathrm{j}}+\frac{7}{4}\hat{\mathrm{k}}.$