Given, l(3a+2b+c)+m(2a+2b+3c)+n(a+2b+5c)=0 Where l,m,n are scalar ∴a(3l+2m+n)+2b(l+m+n)+c(l+3m+5n)=0 a,b,c are independent vector ∴3l+2m+n=0 . . . (i) l+m+n=0 . . . (ii) l+3m+5n=0 . . . (iii) From Eqs. (ii) and (iii), m+2n=0 Putting the value of m in Eq. (i), we get 3l+2(−2n)+n=0⇒3l−3n=0⇒1=n l=n,m+2n=0