We know that,

Domain of ${\sec }^{-1}x$ is $(-\infty ,-1]\cup [1,\infty )$.

Now, $f\left(x\right)={\sec }^{-1}\left(\frac{2x}{5x+3}\right)$

So,

$\frac{2x}{5x+3}\le -1 \text{or} \frac{2x}{5x+3}\ge 1$

$\Rightarrow \frac{2x}{5x+3}+1\le 0 \text{or} \frac{2x}{5x+3}-1\ge 0$

$\Rightarrow \frac{7x+3}{5x+3}\le 0 \text{or} \frac{-3x-3}{5x+3}\ge 0$

$\Rightarrow \frac{7x+3}{5x+3}\le 0 \text{or} \frac{x+1}{5x+3}\le 0$

$\Rightarrow x\in [-1,-\frac{3}{5})\cup (-\frac{3}{5},-\frac{3}{7}]$

So,

$\alpha =-1, \beta =-\frac{3}{5}, \gamma =-\frac{3}{5}, \delta =-\frac{3}{7}$

Now,

$$\begin{gathered} \left|3\alpha +10\left(\beta +\gamma \right)+21\delta \right| \\ =\left|-3-6-6-9\right| \\ =24 \end{gathered}$$

Hence this is the required answer.