Official solution
n = 2 (l + m)
lm + n(l + m) = 0
lm + 2(l + m)2 = 0
2l2 + 2m2 + 5ml = 0
\[2{\left( {{l \over m}} \right)^2} + 2 + 5\left( {{l \over m}} \right) = 0\]
2t2 + 5t + 2 = 0
(t + 2)(2t + 1) = 0
\[ \Rightarrow t = - 2; - {1 \over 2}\]
(i) \[{l \over m} = - 2\]
\[{n \over m} = - 2\]
(-2m, m, -2m)
(-2, 1, -2)
(ii) \[{l \over m} = - {1 \over 2}\]
n = -2l
(l, -2l, -2l)
(1, -2, -2)
\[\cos \theta = {{ - 2 - 2 + 4} \over {\sqrt 9 \sqrt 9 }} = 0 \Rightarrow 0 = {\pi \over 2}\]