Find the angle between the lines whose direction cosines are given by the equations `3l + m + 5n = 0` and `6mn - 2nl + 5lm = 0`

Correct Answer - c
`3l+m+5n =0 " "`(i)
`" "6mn-2nl+5ml=0" "`(ii)
Substituting the value of `n` from (i) in (ii), we get
`" "6l^(2)+ 9lm - 6m^(2)=0`
or `" "6((l)/(m))^(2)+ 9((l)/(m))-6=0`
`therefore" "(l_1)/(m_1)= (1)/(2) and (l_2)/(m_2)=-2`
From Eq. (i), we get
`" "(l_1)/(n_1)= -1 and (l_2)/(n_2)= -2`
`therefore" "(l_1)/(1)= (m_1)/(2)= (n_2)/(-1) = sqrt ((l_(1)^(2)+ m_(1)^(2)+ n_(1)^(2))/(1+4+1))= (1)/(sqrt6)`
and `" "(l_2)/(2)= (m_2)/(-1)= (n_2)/(-1) = (sqrt(l_(2)^(2)+ m_(2)^(2)+ n_(2)^(2)))/(sqrt(4+1+1)) = (1)/(sqrt6)`
If `theta` be the angle between the lines, then
`" "costheta= ((1)/(sqrt6))((2)/(sqrt6))+ ((2)/(sqrt6))(-(1)/(sqrt6)) + (-(1)/(sqrt6))(-(1)/(sqrt6))= (1)/(6)`
or `" "theta= cos^(-1)((1)/(6))`