Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

$$\begin{array}{rl}& l-5\mathrm{m}+3\mathrm{n}=0\text{ and }7l^2+5{\mathrm{m}}^2-3{\mathrm{n}}^2=0\\ & \Rightarrow l=5\mathrm{m}-3\mathrm{n}\text{ and }7l^2=3{\mathrm{n}}^2-5{\mathrm{m}}^2\end{array}$$ Putting $l=(5\mathrm{m}-3\mathrm{n})$ in $7l^2=3{\mathrm{n}}^2-5{\mathrm{m}}^2$, we get $$\begin{array}{rl}& 7(5m-3n{)}^2=3n^2-5m^2\\ & \Rightarrow 7\left(25m^2-30mn+9n^2\right)=3n^2-5m^2\\ & \Rightarrow 180m^2-210mn+60n^2=0\\ & \Rightarrow 6m^2-7mn+2n^2=0\\ & \Rightarrow (3m-2h)(2m-n)=0\\ & \Rightarrow 3m=2n\text{ or }2m=n\end{array}$$ If $3\mathrm{m}=2\mathrm{n}$, then $l=\frac{\mathrm{n}}{3}$ $$\therefore \frac{\mathrm{m}}{2}=\frac{\mathrm{n}}{3}=\frac{l}{1}=\frac{1}{\sqrt{14}}$$ $$\therefore l+\mathrm{m}+\mathrm{n}=\frac{6}{\sqrt{14}}$$ If $2\mathrm{m}=\mathrm{n}$, then $l=\frac{-\mathrm{n}}{2}$ $$\therefore \frac{\mathrm{m}}{1}=\frac{\mathrm{n}}{2}=\frac{l}{-1}=\frac{1}{\sqrt{6}}$$ $$\therefore l+\mathrm{m}+\mathrm{n}=\frac{2}{\sqrt{6}}$$ $\therefore$ The possible values of $l+\mathrm{m}+\mathrm{n}$ is $\frac{2}{\sqrt{6}}$ or $\frac{6}{\sqrt{14}}$