Answer:
Option B
Explanation:
We have,
$a=x\hat{i}+y\hat{j}+z\hat{k} $
$(a \times \hat{i})(\hat{i}+\hat{j})+(\hat{a} \times \hat{j}) (\hat{j}+\hat{k})+(a \times \hat{k})(\hat{k} +\hat{i})$
=$[a \hat{i}\hat{i}]+[a\hat{i}\hat{j}]+[a \hat{j}\hat{j}]+[a \hat{j} \hat{k}]+[a \hat{k} \hat{k}]+[a \hat{k}\hat{i}]$
=$\left[a \hat{i}\hat{j}\right]+\left[a \hat{j} \hat{k}\right]+\left[a \hat{k} \hat{i}\right]$
$[\because [a \hat{i}\hat{i}]=[a \hat{j} \hat{j}]=[a \hat{k} \hat{k}]=0]$
=$a.(\hat{i} \times \hat{j})+a.(\hat{j} \times \hat{k})+a.(\hat{k} \times \hat{i})$
= $a.\hat{k}+a.\hat{i}+a.\hat{j}=a.(\hat{i}+\hat{j}+\hat{k})$
=$(x \hat{i}+y\hat{j}+z \hat{k}).(\hat{i}+\hat{j}+\hat{k})=x+y+z$
