- A$\sqrt \lambda + \sqrt \mu = 1$
- B$\lambda \ne \mu $
- C$\lambda + \mu = 0$
- ✓$\lambda = \mu $
$x=\sqrt{\lambda} y+(\sqrt{\lambda}-1) \Rightarrow y=\frac{x-(\sqrt{\lambda}-1)}{\sqrt{\lambda}}$ .....$(i)$
$z=(\sqrt{\lambda}-1) y+\sqrt{\lambda} \Rightarrow y=\frac{z-\sqrt{\lambda}}{\sqrt{\lambda}-1}$ ......$(ii)$
From $(i)$ and $(ii)$
$\frac{x-(\sqrt{\lambda}-1)}{\sqrt{\lambda}}=\frac{y-0}{1}=\frac{z-\sqrt{\lambda}}{\sqrt{\lambda}-1}$ ....$(A)$
The equation $(\mathrm{A})$ is the equation of line $\mathrm{L}_{1}$
Similarly equation ofline $\mathrm{L}_{2}$ is
$x-\frac{(1-\sqrt{\mu})}{\sqrt{\mu}}=\frac{y-0}{1}=\frac{z-\sqrt{\mu}}{1-\sqrt{\mu}}$ ....$(B)$
Since $\mathrm{L}_{1} \perp \mathrm{L}_{2},$ therefore
$\sqrt{\lambda} \sqrt{\mu}+1 \times 1+(\sqrt{\lambda}-1)(1-\sqrt{\mu})=0$
$\Rightarrow \sqrt{\lambda}+\sqrt{\mu}=0 \Rightarrow \sqrt{\lambda}=-\sqrt{\mu}$
$\Rightarrow \lambda=\mu$
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