For any two vectors $\vec{a}$ and $\vec{b}$, the expression $(\vec{a} \times \hat{i}) .(\vec{b} \times \hat{i})+(\vec{a} \times \hat{j}) .(\vec{b} \times \hat{j})+(\vec{a} \times \hat{k}),(\vec{b} \times \hat{k})$ is always equal to:

$2 \vec{a} . \vec{b}$

$(\vec{a} \times \hat{i}) .(\vec{b} \times \hat{i})=\left|\begin{array}{ll}\vec{a} . \vec{b} & \vec{a} . \hat{i} \\ \vec{b} . \hat{i} & \hat{i} . \hat{i}\end{array}\right|$

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

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

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

Let  $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}, \vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}$

$\Rightarrow(\vec{a} . \hat{i})=a_1,(\vec{a} . \hat{j})=a_2,(\vec{a} . \hat{k})=a_3$

$(\vec{b} . \hat{i})=b_1,(\vec{b} . \hat{j})=b_2,(\vec{b} . \hat{k})=b_3$

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

$=3 \vec{a} . \vec{b}-\left(a_1 b_1+a_2 b_2+a_3 b_3\right)$

$=3 \vec{a} . \vec{b}-\vec{a} . \vec{b}$

$=2 \vec{a} . \vec{b}$

Hence (2) is correct answer.