Given,$f\left(x\right)=2x^3+(2p-7)x^2+3(2p-9)x-6$

On taking differentiation w.r.t $\mathrm{x}$, we get

$f^{\text{'}}\left(x\right)=6x^2+2(2p-7)x+3(2p-9)$

$\Rightarrow f^{\text{'}}\left(0\right)=6(0{)}^2+2(2p-7\left)\right(0)+3(2p-9)$

$\Rightarrow f^{\text{'}}\left(0\right)=3(2p-9)$

Here, given that for $x<0$ the function has maxima and for $x>0$ the function has minima so the function is decreasing function about $x=0$,

Hence $f^{\text{'}}\left(0\right)<0$

So, $3(2p-9)<0$

$\Rightarrow \mathrm{p}<\frac{9}{2}$

Therefore, $\mathrm{p}\in \left(-\infty ,\frac{9}{2}\right)$