## University Calculus: Early Transcendentals (3rd Edition)

$\lt 0,-3,3 \gt$
Since, we have $n=\lt 2,1,-1 \gt$ and $v=\lt 1,1,1 \gt$ Thus, the cross product is: $n \times v=i(-1+1)-j(2+1)+k(2+1)=-3j+3k$ Hence, we have $\lt 0,-3,3 \gt$