TS EAMCET 14-Sep-2020 Shift 2 Solved Paper

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