TS EAMCET 5 May 2018 Shift 2 Solved Paper

It is given that,
l+3m+5n=0
l=−(3m+5n) …… (I)
Thus,
5(−3m−5n)m−2mn +6(−3m−5n)n=0
−15m2−25mn−2mn−18mn −30n2=0
15m2+45mn+30n2=0
m2+3mn+2n2=0
Solve further,
(m+n)(m+2n)=0
m=−n,−2n
When m=−n, then l=−2n.
Thus, the DR 'S are 2,-1,1 .
When m=−2n, then l=n .
Thus, the DR'S are 1,1,-2 .
Now,
=

1

√6√6

θ=cos−1(

1

6

)