If the direction cosines( l,m,n )of a line are related by the equation $l+\mathrm{m}+\mathrm{n}=0, 2\mathrm{mn}+2ml-nl=0$. Then the ordered triplet (l,m,n) is

$\begin{array}{l}\mathrm{l}+\mathrm{m}+\mathrm{n}=0\text{ and }2\mathrm{mn}+2\mathrm{ml}-\mathrm{nl}=0\\ \mathrm{l}=-(\mathrm{m}+\mathrm{n})\Rightarrow 2\mathrm{mn}-(2\mathrm{m}-\mathrm{n})(\mathrm{m}+\mathrm{n})=0\\ \Rightarrow 2{\mathrm{m}}^2-{\mathrm{n}}^2-\mathrm{mn}=0\Rightarrow \frac{\mathrm{m}}{\mathrm{n}}=1\text{ or }\frac{-1}{2}\end{array}$
When $\mathrm{m}=\mathrm{n},\mathrm{l}=-2\mathrm{n};\text{ when }\mathrm{m}=\frac{-\mathrm{n}}{2},\mathrm{l}=\frac{-\mathrm{n}}{2}$
Hence $$\begin{gathered} \frac{l}{-2}=\frac{\mathrm{m}}{1}=\frac{\mathrm{n}}{1}\text{ or }\frac{\mathrm{l}}{-1}=\frac{\mathrm{m}}{-1}=\frac{\mathrm{n}}{2} \\ d.r\text{'}s of lines are (-2,1,1) and (-1,-1,2) \\ direction \cos ines of lines are (\frac{-2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}}) , (\frac{-1}{\sqrt{6}},\frac{-1}{\sqrt{6}},\frac{2}{\sqrt{6}}) \end{gathered}$$