Let $f\left(x\right)=\left\{\begin{array}{l}x-1, x \text{is even},\\ 2x, x \text{is odd},\end{array}\right.x\in N$. If for some $a\in N,f\left(f\left(f\left(a\right)\right)\right)=21$, then $\underset{x\to a^{-}}{\lim }\left\{\frac{{\left|x\right|}^3}{a}-\left[\frac{x}{a}\right]\right\},$ where $\left[t\right]$ denotes the greatest integer less than or equal to $t$, is equal to:

Given: $f\left(x\right)=\left\{\begin{array}{l}x-1; x \text{is even}\\ 2x: x \text{is odd}\end{array}\right.$

And $f\left(f\left(f\left(a\right)\right)\right)=21$

Case-I: If $a$ is even

$\Rightarrow f\left(a\right)=a-1=\text{Odd}$

$\Rightarrow f\left(f\left(a\right)\right)=2\left(a-1\right)=\text{Even}$

$\Rightarrow f\left(f\left(f\left(a\right)\right)\right)=2a-3=21$

$\Rightarrow a=12$

Case-II: If $a$ is odd

$\Rightarrow f\left(a\right)=2a=\text{Even}$

$\Rightarrow f\left(f\left(a\right)\right)=2a-1=\text{Odd}$

$\Rightarrow f\left(f\left(f\left(a\right)\right)\right)=4a-2=21$ (Not possible)

Hence, $a=12$

Now,$\underset{x\to {12}^{-}}{\lim }\left(\frac{{\left|x\right|}^3}{12}-\left[\frac{x}{12}\right]\right)=\underset{x\to {12}^{-}}{\lim }\frac{{\left|x\right|}^3}{12}-\underset{x\to {12}^{-}}{\lim }\left[\frac{x}{12}\right]$

$\Rightarrow \underset{x\to {12}^{-}}{\lim }\left(\frac{{\left|x\right|}^3}{12}-\left[\frac{x}{12}\right]\right)=144-0$

$\Rightarrow \underset{x\to {12}^{-}}{\lim }\left(\frac{{\left|x\right|}^3}{12}-\left[\frac{x}{12}\right]\right)=144$