(a) A(1,3,5),B(2,4,6),C(4,5,k) ⇒AB=(1,1,1);BC=(2,1,k−6) AC=(3,2,k−5) According to the question, AB⋅BC=0 ⇒2+1+k−6=0⇒k=3 AB⋅AC=0 ⇒3+2+k−5=0⇒k=0 BC⋅AC=0 ⇒6+2+(k−6)(k−5)=0⇒k2−11k+38=0 ∴D<0 ⇒ No real values of k Hence, number of possible values of k is 2 .