Show that the lines whose D.C’s are given by  l+ m + n = 0 ,2mn + 3nl - 5lm = 0 are perpendicular to each other.

Given eqs are l + m + n = 0 ...(1)
$2\mathrm{mn}+3\mathrm{nl}-5\mathrm{lm}=0 ...\left(2\right)$
Case (i) : From eq (1), m = –l – n
Substituting m = – l – n in eq (2)
$\begin{array}{l}\Rightarrow 2(-\mathrm{l}-\mathrm{n})\mathrm{n}+3\mathrm{nl}-5\mathrm{l}(-\mathrm{l}-\mathrm{n})=0\\ \Rightarrow -2\ln -2{\mathrm{n}}^2+3\mathrm{nl}+5{\mathrm{l}}^2+5\ln =0\\ \Rightarrow 5{\mathrm{l}}^2+6\ln -2{\mathrm{n}}^2=0\end{array}$
$$\begin{gathered} \text{ dividing both sides by }{\mathrm{n}}^2 \\ \Rightarrow 5\frac{{\mathrm{l}}^2}{{\mathrm{n}}^2}+6\frac{\mathrm{l}}{\mathrm{n}}-2=0\to \left(3\right) \end{gathered}$$
$\text{ Let roots of equation (3) are }\frac{{\mathrm{l}}_1}{{\mathrm{n}}_1},\frac{{\mathrm{l}}_2}{{\mathrm{n}}_2}$
$\text{ Product of roots }=\frac{\mathrm{c}}{\mathrm{a}}$
$\Rightarrow \frac{{\mathrm{l}}_1}{{\mathrm{n}}_1}\cdot \frac{{\mathrm{l}}_2}{{\mathrm{n}}_2}=\frac{-2}{5}\Rightarrow \frac{{\mathrm{l}}_1{\mathrm{l}}_2}{-2}=\frac{{\mathrm{n}}_1{\mathrm{n}}_2}{5}$
Case (ii): From eq (1) l = – m – n
Substituting l = – m – n in eq (2)
$\begin{array}{l}\Rightarrow 2\mathrm{mn}+3\mathrm{n}(-\mathrm{m}-\mathrm{n})-5(-\mathrm{m}-\mathrm{n})\mathrm{m}=0\\ \Rightarrow 2\mathrm{mn}-3\mathrm{mn}-3{\mathrm{n}}^2+5{\mathrm{m}}^2+5\mathrm{mn}=0\\ \Rightarrow 5{\mathrm{m}}^2+4\mathrm{mn}-3{\mathrm{n}}^2=0\\ \text{ dividing both sides by }{\mathrm{n}}^2\\ \Rightarrow 5\frac{{\mathrm{m}}^2}{{\mathrm{n}}^2}+4\frac{\mathrm{m}}{\mathrm{n}}-3=0\dots \text{ (4) }\end{array}$
$\text{ Let }\frac{{\mathrm{m}}_1}{{\mathrm{n}}_1},\frac{{\mathrm{m}}_2}{{\mathrm{n}}_2}\text{ are roots of eq (4) }$
$\text{ Product of roots }=\frac{\mathrm{c}}{\mathrm{a}}$
$\begin{array}{r}\Rightarrow \frac{{\mathrm{m}}_1}{{\mathrm{n}}_1}\cdot \frac{{\mathrm{m}}_2}{{\mathrm{n}}_2}=\frac{-3}{5}\\ \Rightarrow \frac{{\mathrm{m}}_1{\mathrm{m}}_2}{-3}=\frac{{\mathrm{n}}_1{\mathrm{n}}_2}{5}\end{array}$
From case (i) & (ii)
$\begin{array}{l}\frac{{\mathrm{l}}_1{\mathrm{l}}_2}{-2}=\frac{{\mathrm{m}}_1{\mathrm{m}}_2}{-3}=\frac{{\mathrm{n}}_1{\mathrm{n}}_2}{5}=\mathrm{k}\\ {\mathrm{l}}_1{\mathrm{l}}_2=-2\mathrm{k};{\mathrm{m}}_1{\mathrm{m}}_2=-3\mathrm{k};{\mathrm{n}}_1{\mathrm{n}}_2=5\mathrm{k}\\ {\mathrm{l}}_1{\mathrm{l}}_2+{\mathrm{m}}_1{\mathrm{m}}_2+{\mathrm{n}}_1{\mathrm{n}}_2=-2\mathrm{k}-3\mathrm{k}+5\mathrm{k}\\ =-5\mathrm{k}+5\mathrm{k}=0\end{array}$
$\therefore \text{ Given lines are perpendicular to each other }$