$(\vec{a} . \hat{i})^2+(\vec{a} . \hat{j})^2+(\vec{a} . \vec{k})^2$ is equal to:

1 $\vec{a}$ $-\vec{a}$ $|\vec{a}|^2$

$|\vec{a}|^2$

let  $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$ $|\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2}$ so  $(\vec{a} . \hat{i})^2+(\vec{a} . \hat{j})^2+(\vec{o} . \hat{k})^2$ $=a_1^2+a_2^2+a_3^2$ $=|\vec{a}|^2$