A straight line $L$cuts the lines $AB, AC$ and $AD$of a parallelogram $ABCD$ at points $B_1,C_1,$and $D_1$ respectively. If ${\overrightarrow{\mathrm{AB}}}_1={\mathrm{\lambda }}_1\overrightarrow{\mathrm{AB}},{\overrightarrow{\mathrm{AD}}}_1={\mathrm{\lambda }}_2\overrightarrow{\mathrm{AD}}\text{ and }{\overrightarrow{\mathrm{AC}}}_1={\mathrm{\lambda }}_3\overrightarrow{\mathrm{AC}}$. then $\frac{1}{{\mathrm{\lambda }}_3}-(\frac{1}{{\mathrm{\lambda }}_1}+\frac{1}{{\mathrm{\lambda }}_2})+1=$

Let $\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{a},} \mathrm{and} \overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{b}}$
then $\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$
Given, $\overrightarrow{{\mathrm{AB}}_1}={\mathrm{\lambda }}_1\overrightarrow{\mathrm{a}},\overrightarrow{{\mathrm{AD}}_1}={\mathrm{\lambda }}_2\overrightarrow{\mathrm{b}}$
$$\begin{gathered} \overrightarrow{{\mathrm{AC}}_1}={\mathrm{\lambda }}_3(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \\ \overrightarrow{{\mathrm{B}}_1{\mathrm{D}}_1}=\overrightarrow{{\mathrm{AD}}_1}-\overrightarrow{{\mathrm{AB}}_1} \\ \overrightarrow{{\mathrm{B}}_1{\mathrm{D}}_1}={\mathrm{\lambda }}_2\overrightarrow{\mathrm{b}}-{\mathrm{\lambda }}_1\overrightarrow{\mathrm{a}} \end{gathered}$$
$\because$vectors $\overline{{\mathrm{D}}_1{\mathrm{C}}_1}$ and $\overline{{\mathrm{B}}_1{\mathrm{D}}_1}$ are collinear we have $\overline{{\mathrm{D}}_1{\mathrm{C}}_1}=\mathrm{k}\overline{{\mathrm{B}}_1{\mathrm{D}}_1}$ for some $\mathrm{k}\in \mathrm{R}$
 $\begin{array}{l}\Rightarrow \overrightarrow{{\mathrm{AC}}_1}-\overrightarrow{{\mathrm{AD}}_1}=\mathrm{K}\overrightarrow{{\mathrm{B}}_1{\mathrm{D}}_1}\\ \Rightarrow {\mathrm{\lambda }}_3(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})-{\mathrm{\lambda }}_2\overrightarrow{\mathrm{b}}=\mathrm{k}\left({\mathrm{\lambda }}_2\overrightarrow{\mathrm{b}}-{\mathrm{\lambda }}_1\overrightarrow{\mathrm{a}}\right)\end{array}$

${\mathrm{\lambda }}_3\overrightarrow{\mathrm{a}}+\left({\mathrm{\lambda }}_3-{\mathrm{\lambda }}_2\right)\overrightarrow{\mathrm{b}}={\mathrm{k\lambda }}_2\overrightarrow{\mathrm{b}}-{\mathrm{k\lambda }}_1\overrightarrow{\mathrm{a}}$
Hence, ${\mathrm{\lambda }}_3=-{\mathrm{K\lambda }}_1$
${\mathrm{\lambda }}_3-{\mathrm{\lambda }}_2={\mathrm{K\lambda }}_2$
$\Rightarrow \mathrm{K}=\frac{-{\mathrm{\lambda }}_3}{{\mathrm{\lambda }}_1}=\frac{{\mathrm{\lambda }}_3-{\mathrm{\lambda }}_2}{{\mathrm{\lambda }}_2}$
(OR)
${\mathrm{\lambda }}_1{\mathrm{\lambda }}_2={\mathrm{\lambda }}_1{\mathrm{\lambda }}_3+{\mathrm{\lambda }}_2{\mathrm{\lambda }}_3$ (OR)
$\frac{1}{{\mathrm{\lambda }}_3}=\frac{1}{{\mathrm{\lambda }}_1}+\frac{1}{{\mathrm{\lambda }}_2}$