According to the question, l+m+n=0.....(i) ⇒‌‌l=−(m+n) and l2=m2+n2....(ii) ⇒‌‌{−(m+n)}2=m2+n2 ⇒‌‌m2+n2+2mn=m2+n2 ⇒‌‌2mn=0 ∴‌‌mn=0 Case I.m=0 Then, ‌‌l=−n Let l=k,m=0,n=−k ∴‌‌l2+m2+n2=1 ⇒‌‌k2+0+k2=1 ⇒‌‌2k2=1⇒k=‌
1
√2
Apply, l=‌
1
√2
,m=0,n=‌
−1
√2
Case II.n=0 Then, l=−m ∴‌‌l2+m2+n2=1 ⇒‌‌k2+k2+0=0 ⇒‌‌k=‌
1
√2
apply, l=‌
1
√2
,m=−‌
1
√2
,n=0 Now, Angle between above two lines (cos‌θ)