1) l = m + n
2) m = l + n
3) n = l + m
4) l + m + n = 0
Solution:
Given lx + my + n = 0, mx + ny + l = 0, nx + ly + m = 0 are concurrent.
So
\(\begin{array}{l}\begin{vmatrix} a_{1} & b_{1}&c_{1}\ a_{2}& b_{2} & c_{2}\ a_{3}& b_{3} & c_{3} \end{vmatrix} = 0\end{array} \)
=>
\(\begin{array}{l}\begin{vmatrix} l & m&n\ m& n & l\ n& l & m \end{vmatrix} = 0\end{array} \)
=> l(mn – l2) -m(m2 – nl) +n(ml – n2) = 0
=> 3lmn – (l3 + m3 + n3) = 0
=> l3 + m3 + n3– 3lmn = 0
=> l + m + n = 0
Hence option (4) is the answer.
