Given that a, b and c are non-zero vectors and a+2b is collinear with c and b+3c is collinear with a. ∴a+2b=λc....(i) and b+3c=µa....(ii)  [from same λ,µ] Put the value of b in Eq. (i) by Eq (ii), we get a+2(µa−3c)=λc ⇒a+2µa−6c=λc ⇒a(1+2µ)−c(λ+6)=0  [∵ a and c are non-collinear] ⇒1+2µ=0 and λ+6=0 ⇒λ=−6 and µ=−