If the domain of the function $$\sin ^{-1}\left(\frac{3 x-22}{2 x-19}\right)+\log _{\mathrm{e}}\left(\frac{3 x^2-8 x+5}{x^2-3 x-10}\right)$$ is $(\alpha, \beta]$, then $3 \alpha+10 \beta$ is equal to:

<p>$$\begin{aligned} & \sin ^{-1}\left(\frac{3 x-22}{2 x-19}\right)+\log _e\left(\frac{3 x^2-8 x+5}{x^2-3 x-10}\right) \\ & -1 \leq \frac{3 x-22}{2 x-19} \leq 1 \\ & \frac{3 x-22}{2 x-19}+1 \geq 0 \text { and } \frac{3 x-22}{2 x-19}-1 \leq 0 \\ & \frac{3 x-22+2 x-19}{2 x-19} \geq 0 \text { and } \frac{3 x-22-2 x+19}{2 x-19} \leq 0 \\ & \Rightarrow \frac{5 x-41}{2 x-19} \geq 0 \text { and } \frac{x-3}{2 x-19} \leq 0 \\ & x \in\left(-\infty, \frac{41}{5}\right] \cup\left(\frac{19}{2}, \infty\right) \text { and } x \in\left[3, \frac{19}{2}\right) \end{aligned}$$</p> <p>$$\begin{aligned} & \Rightarrow \quad x \in\left[3, \frac{41}{5}\right] \quad \text{.... (1)}\\ & \text { and, } \frac{3 x^2-8 x+5}{x^2-3 x-10}>0 \\ & \frac{(3 x-5)(x-1)}{(x-5)(x-2)}>0 \\ & \Rightarrow \quad x \in(-\infty,-2) \cup\left[1, \frac{5}{3}\right] \cup(5, \infty) \ldots \end{aligned}$$</p> <p>Taking intersection of individual domains</p> <p>$x \in\left(5, \frac{41}{5}\right]$</p> <p>$$\begin{aligned} & \Rightarrow \quad \alpha=5 \text { and } \beta=\frac{41}{5} \\ & \Rightarrow 3 \alpha+10 \beta=15+82 \\ & =97 \end{aligned}$$</p> <p>$\therefore \quad$ Option (4) is correct</p>