If domain of the function ${\log }_{\mathrm{e}}\left(\frac{6{\mathrm{x}}^2+5\mathrm{x}+1}{2\mathrm{x}-1}\right)+{\cos }^{-1}\left(\frac{2{\mathrm{x}}^2-3\mathrm{x}+4}{3\mathrm{x}-5}\right)\text{ is }(\mathrm{\alpha },\mathrm{\beta })\cup (\mathrm{\gamma },\mathrm{\delta }]$, then $18\left({\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+{\mathrm{\gamma }}^2+{\mathrm{\delta }}^2\right)$ is equal to ……….

${\log }_{\mathrm{e}}\left(\frac{6{\mathrm{x}}^2+5\mathrm{x}+1}{2\mathrm{x}-1}\right)+{\cos }^{-1}\left(\frac{2{\mathrm{x}}^2-3\mathrm{x}+4}{3\mathrm{x}-5}\right)$
$$\begin{gathered} \begin{array}{l}\frac{6{\mathrm{x}}^2+5\mathrm{x}+1}{2\mathrm{x}-1}>0 \\ \Rightarrow (2x+1)(3x+1)(2x-1)>0\\ \Rightarrow \mathrm{x}\in \left(\frac{-1}{2},\frac{-1}{3}\right)\cup \left(\frac{1}{2},\mathrm{\infty }\right).....\left(1\right)\\ -1\le \frac{2{\mathrm{x}}^2-3\mathrm{x}+4}{3\mathrm{x}-5}\le 1\\ \Rightarrow \mathrm{x}\in \left(\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\dots ..\left(2\right) \\ 2{\mathrm{x}}^2-3\mathrm{x}+4<3x-5 \\ 2x^2-6x+9<0 \\ \begin{array}{l}\mathrm{x}\in \left(\frac{-1}{2},\frac{-1}{3}\right)\cup \left(\frac{1}{2},\frac{1}{\sqrt{2}}\right)\\ {\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+{\mathrm{\gamma }}^2+{\mathrm{\delta }}^2=\frac{1}{4}+\frac{1}{9}+\frac{1}{4}+\frac{1}{2}\\ =1+\frac{1}{9}=\frac{10}{9}, \\ 18\left({\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+{\mathrm{\gamma }}^2+{\mathrm{\delta }}^2\right)=18\times \frac{10}{9}=20\end{array}\end{array} \end{gathered}$$