Given that the points with position vectors 60i+3j,40i−8j and i−52j are collinear, then there exist three scalars 1,x and y such that ‌1+x+y=0 . . . (i) and(60i+3j)⋅1+(40i−8j)⋅x +(ai−52j)⋅y=0 ‌⇒‌‌(60+40x+ay)i+(3−8x−52y)j=0 ‌‌=0i+0j
Comparing both sides, we get ‌‌60+40x+ay=0 . . . (ii) and 3−8x−52y=0 . . . (iii) Solving Eqs. (i) and (iii), we get x=‌