Given binomial expansion of $\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^{l}}\right)^{9}$
$\begin{aligned} T_{r+1} & ={ }^{9} C_{r}\left(\frac{x^{\frac{5}{2}}}{2}\right)^{9-r}\left(\frac{-4}{x^{l}}\right)^{r} \\\\ & ={ }^{9} C_{r} x^{\frac{45-5 r}{2}-lr} \cdot 2^{r-9} \cdot 4^{r} \cdot(-1)^{r} \end{aligned}$
For constant term, power of x is zero.
So, $\frac{45-5 r}{2}=l r \Rightarrow 2 l r+5 r=45$
Now constant term $=-84$
and ${ }^{9} C_{r} \cdot 2^{3 r-9}(-1)^{r}=-84$
So, $r=3$ and $l=5$
Now for $x^{-15}, \frac{45-5 r}{2}-5 r=-15$
$\Rightarrow 45-15 r=-30$
$\Rightarrow r=5$
$\therefore$ Coefficient $=-{ }^{9} C_{5} 2^{6}=-63.2^{7}$
$\therefore \alpha=7, \beta=-63$
and $|\alpha l-\beta|=|7 \times 5+63|=98$