Given that $f\left(x\right)=\log \left(4x^2+11x+6\right)+{\sin }^{-1}\left(4x+3\right)+{\cos }^{-1}\left(\frac{10x+6}{3}\right)$

We know that the domain of $A\left(x\right)+B\left(x\right)+C\left(x\right)$ is $D\left(A\right)\cap D\left(B\right)\cap D\left(C\right)$.

Let us find the domain of $\log \left(4x^2+11x+6\right)$.

We know that domain of $\log \left(f\left(x\right)\right)$ is $f\left(x\right)>0$.

$\Rightarrow 4x^2+11x+6>0$

$\Rightarrow 4x^2+8x+3x+6>0$

$\Rightarrow \left(4x+3\right)\left(x+2\right)>0$

$\Rightarrow x\in \left(-\infty ,-2\right)\cup \left(\frac{-3}{4},\infty \right)$

Domain of $\log \left(4x^2+11x+9\right)$ is $x\in \left(-\infty ,-2\right)\cup \left(\frac{-3}{4},\infty \right)$.

Let us find the domain of ${\sin }^{-1}\left(4x+3\right)$

$\Rightarrow -1\le 4x+3\le 1$

$\Rightarrow -1\le x\le \frac{-1}{2}$

$\Rightarrow x\in \left[-1,\frac{-1}{2}\right]$

Let us find the domain of ${\cos }^{-1}\left(\frac{10x+6}{3}\right)$.

$\Rightarrow -1\le \frac{10x+6}{3}\le 1$

$\Rightarrow -3\le 10x+6\le 3$

$\Rightarrow \frac{-9}{10}\le x\le \frac{-3}{10}$.

Now the common domain is $\frac{-3}{4}\le x\le \frac{-1}{2}$.

$\Rightarrow x\in \left[\frac{-3}{4},\frac{-1}{2}\right]$

But given that the domain of $f\left(x\right)$ is $x\in \left[\alpha ,\beta \right]$.

$\Rightarrow \alpha =\frac{-3}{4}, \beta =\frac{-1}{2}$.

$\Rightarrow \left|36\left(\alpha +\beta \right)\right|=\left|36\left(\frac{-3}{4}+\left(\frac{-1}{2}\right)\right)\right|$

$\Rightarrow \left|36\left(\alpha +\beta \right)\right|=\left|36\times \frac{-5}{4}\right|$

$\Rightarrow \left|36\left(\alpha +\beta \right)\right|=45$

Therefore, the required answer is $45$.