The points P, Q, R and S on the sides OA, OB, OC and AB, respectively, of a regular tetrahedron OABC are   coplanar. It is  given  OPOA=13,OQOB=12,OROC=13 and OSAB=λ ,then the value λ is

Let  OA→=a→,OB→=b→ and OC→=c→then  AB→=b→−a→ and OP→=13a→OQ→=12b→,OR→=13c→Since P, Q,, R and S are coplanar, then PS→=αPQ→+βPR→( PS→ can be written as a linear  combination of  PQ→ and  PR→)=α(OQ→−OP→)+β(OR→−OP→)OS→−OP→=−(α+β)a→3+α2b→+β3c→⇒OS→=(1−α−β)a→3+α2b→+β3c→---igiven  OS→=λAB→=λ(b→−a→)---ii From (i) and (ii), β=0,1−α3=−λ and α2=λ⇒ 2λ=1+3λ or  λ=−1