If the domain of the function log₅(18x - x² - 77) is $(\mathrm{\alpha },\mathrm{\beta })$ and the domain of the function
${\log }_{(\mathrm{x}-1)}\left(\frac{2{\mathrm{x}}^2+3\mathrm{x}-2}{{\mathrm{x}}^2-3\mathrm{x}-4}\right)\text{ is }(\mathrm{\gamma },\mathrm{\delta })$, then ${\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+{\mathrm{\gamma }}^2$ is equal to :

$\begin{array}{l}\mathrm{let} {\mathrm{f}}_1\left(\mathrm{x}\right)={\log }_5\left(18\mathrm{x}-{\mathrm{x}}^2-77\right)\\ \therefore 18\mathrm{x}-{\mathrm{x}}^2-77>0\\ \Rightarrow {\mathrm{x}}^2-18\mathrm{x}+77<0\\ \Rightarrow \mathrm{x}\in (7,11) \\ \Rightarrow \mathrm{\alpha }=7,\mathrm{\beta }=11\\ \mathrm{Let} {\mathrm{f}}_2\left(\mathrm{x}\right)={\log }_{(\mathrm{x}-1)}\left(\frac{2{\mathrm{x}}^2+3\mathrm{x}-2}{{\mathrm{x}}^2-3\mathrm{x}-4}\right)\\ \therefore \mathrm{x}-1>0,\mathrm{x}-1\ne 1 and \frac{2{\mathrm{x}}^2+3\mathrm{x}-2}{{\mathrm{x}}^2-3\mathrm{x}-4}>0\\ \mathrm{x}>1,\mathrm{x}\ne 2 and \frac{(2\mathrm{x}-1)(\mathrm{x}+2)}{(\mathrm{x}-4)(\mathrm{x}+1)}>0\\ \mathrm{x}>1,\mathrm{x}\ne 2\end{array}$

$\begin{array}{l}\therefore \mathrm{x}\in (4,\mathrm{\infty })\\ \therefore \mathrm{\gamma }=4\\ \therefore {\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+{\mathrm{\gamma }}^2=49+121+16\end{array}$
                            = 186