Q67 / 125 · BITSAT 2013 · Mathematics
If the lines $\ell \mathrm{x}+\mathrm{my}+\mathrm{n}=0, \mathrm{mx}+\mathrm{ny}+\ell=0$ and
$\mathrm{nx}+\ell \mathrm{y}+\mathrm{m}=0$ are concurrent then
- Correct answerA$\ell+m+n=0$
- B$\ell-m-n=0$
- C$\ell+m-n=0$
- D$m+n-\ell=0$
Solution
Since the lines are concurrent, so
$$
\Rightarrow \begin{array}{l}
\left|\begin{array}{lll}
\ell & \mathrm{m} & \mathrm{n} \\
\mathrm{m} & \mathrm{n} & \ell \\
\mathrm{n} & \ell & \mathrm{m}
\end{array}\right|=0 \Rightarrow 3 \ell \mathrm{mn}-\ell^{3}-\mathrm{m}^{3}-\mathrm{n}^{3}=0 \\
(\ell+\mathrm{m}+\mathrm{n})\left(\ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}-(\mathrm{m}-\mathrm{mn}-\mathrm{n} \ell)=0\right.
\end{array}
$$
$\Rightarrow \quad \ell+\mathrm{m}+\mathrm{n}=0\left[\because \ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}>6 \mathrm{~m}+\mathrm{mn}+\mathrm{n}\right]$
Hence, the correct answer is option: 'A'.