The maximum negative integral value of $b$ for which the point $\left(2b+3,b^2\right)$ lies above the line $3x-4y-a\left(a-2\right)=0, \forall a\in R$ is

Let $\left(x_1,y_2\right)$ lie on the line
$$\begin{gathered} \therefore 3x_1-4y_2-a\left(a-2\right)=0 \\ \Rightarrow 4y_2=3x_1-a\left(a-2\right) \end{gathered}$$
Now, $y_2<y_1\Rightarrow \frac{3x_1-a\left(a-2\right)}{4}<y_1$
$$ Putting $x_1=2b+3, y_1=b^2$, we get, 

$\Rightarrow 3\left(2b+3\right)-a\left(a-2\right)<4b^2.$
$\Rightarrow a^2-2a+4b^2-6b-9>0$

Now $\forall a\in R$, $D<0$

$\Rightarrow 4-4\left(4b^2-6b-9\right)<0$
$$\begin{gathered} \Rightarrow 1-4b^2+6b+9<0 \\ \Rightarrow 4b^2-6b+10>0 \end{gathered}$$
$$\begin{gathered} \Rightarrow 2b^2-3b-5>0 \\ \Rightarrow \left(2b-5\right)\left(b+1\right)>0 \end{gathered}$$
$\Rightarrow b\in \left(-\infty ,-1\right)\cup \left(\frac{5}{2},\infty \right)$

Hence, the maximum negative integer value of $b$ is $-2$