$(a)\;5\times10^{-3}\;(2 \hat{i}+2 \hat{j}-\hat{k})\qquad(b)\;5\times10^{-4}\;(2 \hat{i}+2 \hat{j}-\hat{k})\qquad(c)\;5\times10^{-3}\;(2 \hat{i}-2 \hat{j}+\hat{k})\qquad(d)\;5\times10^{-3}\;(2 \hat{i}+2 \hat{j}+\hat{k})$

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Answer : (a) $\;5\times10^{-3}\;(2 \hat{i}+2 \hat{j}-\hat{k})$

Explanation : $\overrightarrow{F}_{12}=\large\frac{kq_{1}q_{2}\;(\overrightarrow{r}_{1}-\overrightarrow{r}_{2})}{|\overrightarrow{r}_{1}-\overrightarrow{r}_{2}|^3}$

$\overrightarrow{F}_{12}=\large\frac{9\times10^9\times3\times5\times10^{-12}\;(2 \hat{i}+2 \hat{j}-\hat{k})}{27}$

$\overrightarrow{F}_{12}=10^{-2} \hat{i} + 10^{-2} \hat{j}-5\times10^{-3}\;\hat{k}\;.$

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