Given,

Origin be the centre, $\left(\pm 3,0\right)$ be the foci and $\frac{3}{2}$ be the eccentricity of a hyperbola,

Now we know that foci is $\left(\pm ae,0\right)\equiv \left(\pm 3,0\right)$,

So $ae=3$ and given $e=\frac{3}{2}$ then $a=\frac{1}{2}$,

Now by formula $b^2=a^2\left(e^2-1\right)$ we get $b=\frac{\sqrt{5}}{4}$

So, equation of hyperbola will be $\frac{x^2}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}-\frac{y^2}{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$16$}\right.}}=1$

$\Rightarrow 20x^2-y^2=80$

Now putting the value of $y$ from the line $2x-y-1=0$ in hyperbola to check intersecting or not,

So, $20x^2-{\left(2x-1\right)}^2=80$

$\Rightarrow 16x^2+4x-81=0$

Now discriminant $D=4^2+4\times 16\times 81>0$, hence line does not intersect the hyperbola.