If $\frac{1}{\cos {290}^0}+\frac{1}{\sqrt{3}\sin {250}^0}=\mathrm{\lambda }$then the value of $9{\mathrm{\lambda }}^4+81{\mathrm{\lambda }}^2+97$must be ______

Here,$\cos {290}^{\circ }=\cos \left({270}^{\circ }+{20}^{\circ }\right)=\sin {20}^{\circ }$ and 
$\sin {250}^{\circ }=\sin \left({270}^{\circ }-{20}^{\circ }\right)=-\cos {20}^{\circ }$
$\therefore$The given expression
$\begin{array}{l}=\frac{1}{\sin {20}^0}-\frac{1}{\sqrt{3}\cos {120}^0}=\mathrm{\lambda }\\ \Rightarrow \frac{1}{\sin {20}^0}-\frac{\cos {60}^0}{\sin {60}^0\cos {20}^0}=\mathrm{\lambda }\\ \Rightarrow \frac{\sin {60}^0\cos {20}^0-\cos {60}^0\sin {20}^0}{\sin {20}^0\cos {20}^0\sin {60}^0}=\mathrm{\lambda }\\ \Rightarrow \frac{\sin \left({60}^0-{20}^0\right)}{\sin {40}^0}=\frac{\sqrt{3}}{2}=\mathrm{\lambda }\\ \therefore \mathrm{\lambda }=\frac{4}{\sqrt{3}}\Rightarrow {\mathrm{\lambda }}^2=\frac{16}{3}\end{array}$
The $9{\mathrm{\lambda }}^4+81{\mathrm{\lambda }}^2+97=9\times \frac{256}{9}+81\times \frac{16}{3}+97$
=256+432+97=785