Given $f\left(x\right)=2x^3-9x^2+12x+5, x\in \left[0,3\right]$

Differentiating w.r.t $x$,

$\Rightarrow f^{\prime }\left(x\right)=6x^2-18x+12$

$\Rightarrow f^{\prime }\left(x\right)=6\left(x-1\right)\left(x-2\right)$

For maxima/minima $f^{\text{'}}\left(x\right)=0$

$\Rightarrow 6\left(x-1\right)\left(x-2\right)=0\Rightarrow x=1,2$

Now, $f^{"}\left(x\right)=12x-18$

So at $x=1, f^{"}\left(x\right)<0$

and at $x=2, f^{"}\left(x\right)>0$

So, for absolute maxima/minima

$\Rightarrow f\left(1\right)=10$

$\Rightarrow f\left(2\right)=9$

$\Rightarrow f\left(0\right)=5$

$\Rightarrow f\left(3\right)=14$

$\Rightarrow M=14$

$\Rightarrow m=5$

$\therefore$$M-m=9$