If the points A,B and C have position vectors (2,1,1), (6,-1,2) and (14,-5,P) respectively and if they are collinear, then P =
4
→OA=2^i+^j+^k,→OB=6^i−^j+2^k,→OC=14^i−5^j+P^k⇒→AB=→OB−→OA=4^i−2^j+^k,→AC=→OC−→OA=12^i−6^j+(p−1)^k
A, B, C are collinear ⇒→AC=λ→AB⇒12^i−6^j+(p−1)^k=λ(4^i−2^j+^k)
⇒λ=3,p−1=3⇒p=4.