The ratio of the sixth term from the beginning to the seventh term from the end in the binomial expansion of ${\left(\sqrt[4]{7}+\frac{1}{\sqrt[7]{4}}\right)}^{\mathrm{n}}$, in the increasing powers of $\sqrt[4]{7}$ be 1:1. If the third term from the beginning is $\frac{\mathrm{\lambda }\sqrt{7}}{\sqrt[7]{4^9}},$, then $\mathrm{\lambda }$ is equal to

In ${\left(\frac{1}{\sqrt[7]{4}}+\sqrt[4]{7}\right)}^{\mathrm{n}}$, The general term is  ${\mathrm{T}}_{\mathrm{r}+1}{=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{\left(\frac{1}{\sqrt[7]{4}}\right)}^{\mathrm{n}-\mathrm{r}}(\sqrt[4]{7}{)}^{\mathrm{r}}.$
Given that ${\mathrm{T}}_6:{\mathrm{T}}_{\mathrm{n}-7+2}={\mathrm{T}}_6:{\mathrm{T}}_{\mathrm{n}-5}{=}^{\mathrm{n}}{\mathrm{C}}_5{\left(\frac{1}{\sqrt[7]{4}}\right)}^{\mathrm{n}-5}\left(\sqrt[4]{7}{)}^5{:}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-6}{\left(\frac{1}{\sqrt[7]{4}}\right)}^6\right(\sqrt[4]{7}{)}^{\mathrm{n}-6}$
$\begin{array}{l}{=}^{\mathrm{n}}{\mathrm{C}}_5{\left(\frac{1}{4}\right)}^{\frac{\mathrm{n}-5}{7}}{7}^{\frac{5}{4}}{:}^{\mathrm{n}}{\mathrm{C}}_6{\left(\frac{1}{4}\right)}^{\frac{6}{7}}{7}^{\frac{\mathrm{n}-6}{4}}=1:1\\ \Rightarrow \frac{{ }^{\mathrm{n}}{\mathrm{C}}_5}{{ }^{\mathrm{n}}{\mathrm{C}}_6}{\left(\frac{1}{4}\right)}^{\frac{\mathrm{n}-11}{7}}{7}^{\frac{11-\mathrm{n}}{4}}=\frac{(\mathrm{n}-6)!6!}{(\mathrm{n}-5)!5!}(28{)}^{\frac{11-\mathrm{n}}{4}}=\frac{6}{\mathrm{n}-5}(28{)}^{\frac{11-\mathrm{n}}{4}}=1\Rightarrow \mathrm{n}=11.\end{array}$
Now the third term from the beginning is
${\mathrm{T}}_3={\mathrm{T}}_{2+1}{=}^{11}{\mathrm{C}}_2{\left(\frac{1}{\sqrt[7]{4}}\right)}^{11-2}(\sqrt[4]{7}{)}^2=55\times {\left(\frac{1}{4}\right)}^{\frac{9}{7}}\times \sqrt{7}=\frac{55\times \sqrt{7}}{4^{\frac{9}{7}}}=\frac{\mathrm{\lambda }\sqrt{7}}{\sqrt[7]{4^9}}\Rightarrow \mathrm{\lambda }=55$