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 $A{B}_1={\lambda }_{1}AB,A{D}_1={\lambda }_{2}AD$ and $A{C}_1={\lambda }_{3}AC,$ then $\frac{1}{{\lambda }_3}$ is equal to

$\text{Let}\overline{\mathrm{AB}}=\overline{\mathrm{a}},\overline{\mathrm{AD}}=\overline{\mathrm{b}}\text{, then}\overline{\mathrm{AC}}=\overline{\mathrm{a}}+\overline{\mathrm{b}}\text{.}$

 $\mathrm{Given},\overline{ A{B}_1}={\lambda }_1\overline{a},\overline{A{D}_1}={\lambda }_2\overline{b},\overline{A{C}_1}={\lambda }_3(\overline{a}+\overline{b})$

 $$\begin{gathered} \overline{\mathrm{AC}}=\overline{\mathrm{a}}+\overline{\mathrm{b}}\text{.} \\ \Rightarrow \overline{\mathrm{AC}}=\overline{AB}+\overline{AD} \\ \Rightarrow \frac{1}{{\lambda }_3}\overline{{\mathrm{AC}}_1}=\frac{1}{{\lambda }_1}\overline{A{B}_1}+\frac{1}{{\lambda }_2}\overline{A{D}_1} \\ \Rightarrow \frac{1}{{\lambda }_1}\overline{A{B}_1}+\frac{1}{{\lambda }_2}\overline{A{D}_1}+\left(-\frac{1}{{\lambda }_3}\overline{{\mathrm{AC}}_1}\right)=0 \\ {B}_1,C_1,D_1 are collinear \\ \Rightarrow \frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_3}=0 \\ \therefore \frac{1}{{\lambda }_3}=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2} \\ \boxed{If l\overline{a}+m\overline{b}+n\overline{c}=\overline{O} such that l+m+n=0 then \overline{a},\overline{b},\overline{c} are collinear points} \end{gathered}$$