Given that $f(x)=\sec ^{-1}\left(\frac{2 x}{5 x+3}\right)$ Since, the domain for $\sec ^{-1} x$ is $|x| \geq 1$ Therefore $\left|\frac{2 x}{5 x+3}\right| \geq 1$ $\begin{aligned} & \frac{2 x}{5 x+3} \leq-1 \text { or } \frac{2 x}{5 x+3} \geq 1 \\\\ & \frac{2 x+5 x+3}{5 x+3} \leq 0 \text { or } \frac{2 x-5 x-3}{5 x+3} \geq 0 \\\\ & \frac{7 x+3}{5 x+3} \leq 0 \text { or } \frac{-3(x+1)}{5 x+3} \geq 0 \end{aligned}$ Case I : $7 x+3 \leq 0$ and $5 x+3>0$ $\begin{array}{rlrl} & x \leq-\frac{3}{7} \text { or } x > -\frac{3}{5} \\\\ & \Rightarrow -\frac{3}{5} < x \leq-\frac{3}{7} \end{array}$ Case II : $7 x+3 \geq 0$ and $5 x+3<0$ $x \geq-\frac{3}{7} \text { and } x<-\frac{3}{5}$ Which is not possible Case III : $x+1 \geq 0$ and $5 x+3<0$ $\begin{aligned} & x \geq-1 \text { and } x<-\frac{3}{5} \\\\ & \Rightarrow -1 \leq x<-\frac{3}{5} \end{aligned}$ Case IV : $x+1 \leq 0$ and $5 x+3 \geq 0$ $x \leq-1 \text { and } x \geq-\frac{3}{5}$ Which is not possible $\therefore$ Domain is $\left[-1,-\frac{3}{5}\right) \cup\left(-\frac{3}{5},-\frac{3}{7}\right]$ $\therefore \alpha=-1, \beta=-\frac{3}{5}, \gamma=-\frac{3}{5}, \delta=-\frac{3}{7}$ $\therefore$ $|3 \alpha+10(\beta+\gamma)+21 \delta| =\left|-3+10\left(-\frac{3}{5}-\frac{3}{5}\right)+21\left(-\frac{3}{7}\right)\right|=24$