We need to find the domain of the function ${\log }_e\left(\frac{6x^2+5x+1}{2x-1}\right)+{\cos }^{-1}\left(\frac{2x^2-3x+4}{3x-5}\right)$.

$\Rightarrow \frac{6x^2+5x+1}{2x-1}>0 .....\left(1\right)$ and

$\Rightarrow -1\le \frac{2x^2-3x+4}{3x-5}\le 1......\left(2\right)$
Now From $\left(1\right)$,
$\Rightarrow \frac{6x^2+5x+1}{2x-1}>0$

$\Rightarrow \frac{(3x+1)(2x+1)}{(2x-1)}>0$

Now we get the common region here as
$\Rightarrow x\in \left(-\frac{1}{2},-\frac{1}{3}\right)\cup \left(\frac{1}{2},\infty \right).......\left(\mathrm{a}\right)$

From $\left(2\right)$ we get that
$\Rightarrow -1\le \frac{2x^2-3x+4}{3x-5}\le 1$

$\Rightarrow \frac{2x^2-3x+4}{3x-5}\ge -1......\left(3\right)$

and $\frac{2x^2-3x+4}{3x-5}\le 1.....\left(4\right)$

From $\left(3\right)$ we get
$\Rightarrow \frac{2x^2-3x+4}{3x-5}+1\ge 0$

$\Rightarrow x\in \left[\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right]\cup \left(\frac{5}{3},\infty \right).....\left(\mathrm{b}\right)$

From $\left(4\right)$ we get
$\Rightarrow \frac{2x^2-3x+4}{3x-5}-1\le 0$

$\Rightarrow \frac{2x^2-6x+9}{3x-5}\le 0$

As $D={\left(-6\right)}^2-4\left(2\right)\left(9\right)<0 \text{and} 2>0$, we observe that $2x^2-6x+9>0 \forall x\in \mathrm{ℝ}$
$\Rightarrow \frac{1}{3x-5}\le 0 \left(\because 2x^2-6x+9>0\forall x\in R\right)$

$\Rightarrow x\in \left(-\infty ,\frac{5}{3}\right) ......\left(\mathrm{c}\right)$

$\Rightarrow$ Intersection of $\left(\mathrm{a}\right), \left(\mathrm{b}\right)$ and $\left(\mathrm{c}\right)$ gives us
$x\in \left(-\frac{1}{2},-\frac{1}{3}\right)\cup \left(\frac{1}{2},\frac{1}{\sqrt{2}}\right)$.

On comparing this with $\left(\alpha ,\beta \right)\cup \left(\gamma ,\delta \right)$,
$\Rightarrow {\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2=\frac{10}{9}$
$\Rightarrow 18\left({\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2\right)=20$.

Hence this is the required answer.