$\hat{i} \times(\vec{a} \times \hat{i})+\hat{j} \times(\vec{a} \times \hat{j})+\hat{k} \times(\vec{a} \times \hat{k})$ is always equal to:

$2 \vec{a}$

$\hat{i} \times(\vec{a} \times \hat{i})=(\hat{i} . \hat{i}) \vec{a}-(\vec{a} . \hat{i}) \vec{i}=\vec{a}-(\vec{a} . \hat{i}) \hat{i}$

Similarly,  $\hat{j} \times(\vec{a} \times \hat{j})=\vec{a}-(\vec{a} . \hat{j}) \hat{j}$

and  $\hat{k} \times(\vec{a} \times \hat{k})=\vec{a}-(\vec{a} . \hat{k}) \hat{k}$

$\Rightarrow \hat{i} \times(\vec{a} \times \hat{i})+\hat{j} \times(\vec{a} \times \hat{j})+\hat{k} \times(\vec{a} \times \hat{k})$

$=3 \vec{a}-((\vec{a} . \hat{i}) \hat{i}+(\vec{a} . \hat{j}) \hat{j}+(\vec{a} . \hat{k}) \hat{k})$

$=2 \vec{a}$

Hence (3) is correct answer.