$f\left(x\right)=x^3-3x^2-\frac{3f^{"}\left(2\right)}{2}x+f^{"}\left(1\right) \dots \left(\mathrm{i}\right)$

$\Rightarrow f^{\text{'}}\left(x\right)=3x^2-6x-\frac{3}{2}{f}^{"}\left(2\right) \dots \left(\mathrm{ii}\right)$

$\Rightarrow f^{"}\left(x\right)=6x-6 \dots \left(\mathrm{iii}\right)$

Then,

$f^{"}\left(2\right)=12-6=6$

$f^{"}\left(1\right)=0$

Now,

$f^{\text{'}}\left(x\right)=3x^2-6x-\frac{3}{2}{f}^{"}\left(2\right)$

$\Rightarrow f^{\text{'}}\left(x\right)=3x^2-6x-\frac{3}{2}\times 6$

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

For maxima/minima, we have

$f^{\text{'}}\left(x\right)=0$

$\Rightarrow 3x^2-6x-9=0$

$\Rightarrow x^2-2x-3=0$

$\Rightarrow x^2-3x+x-3=0$

$\Rightarrow x=-1$ and $3$

Now,

$f^{"}\left(x\right)=6x-6$

$f^{"}(-1)=-12<0$ (maxima)

$f^{"}\left(3\right)=12>0$ (minima)

Again

$f\left(x\right)=x^3-3x^2-\frac{3f^{"}\left(2\right)}{2}x+f^{"}\left(1\right)$

$\Rightarrow f\left(x\right)=x^3-3x^2-\frac{3}{2}\times 6x+0$

$\Rightarrow f\left(x\right)=x^3-3x^2-9x$

$\Rightarrow f\left(3\right)=27-27-9\times 3=-27$