Prove that `hat(i)xx(vec(a)xxhat(i))+hat(j)xx(vec(a)xxhat(j))+hat(k)+xx(vec(a)xxhat(k))=2vec(a)` .

Let vec(a)=hat(i)-hat(j), vec(b)=hat(j)-hat(k), vec(c)=hat(k)-hat(i) . If vec(d) is a unit vectors such that vec(a)*vec(d)=0=[vec(b)vec(c)vec(d)] , then vec(d) is (are) :