‌A(0,1,2),B(2,−1,3),C(1,−3,1) ‌∵‌‌BA=<−2,2,−1≥ ‌‌‌BC=<−1,−2,−2> ‌∵‌‌BA⋅BC=(−2)(−1)+2(−2) ‌‌‌+(−1)(−2)=0 So, △ABC is right angled at B. ∴ Orthocenter (H) is at point B=(2,−1,3) and circumcenter (O)=(‌
