If $\overrightarrow{r}=x_1\hat{a}_{x_1}+x_2\hat{a}_{x_2}+x_3\hat{a}_{x_3}$ and $|\overrightarrow{r}|=r$ then $div(r^2\nabla(In\;r))$ is

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If $\overrightarrow{r}=x_1\hat{a}_{x_1}+x_2\hat{a}_{x_2}+x_3\hat{a}_{x_3}$  and  $|\overrightarrow{r}|=r$  then  $div(r^2\nabla(In\;r))$ is

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The correct answer is(B): 3

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If \(\overrightarrow{r}=x_1\hat{a}_{x_1}+x_2\hat{a}_{x_2}+x_3\hat{a}_{x_3}\) and \(|\overrightarrow{r}|=r\) then \(div(r^2\nabla(In\;r))\)is