Let, $f\left(x\right)=x^3+a{x}^2+bx+c$

$\Rightarrow f\text{'}\left(x\right)=3x^2+2ax+b$

$\Rightarrow f\text{'}\text{'}\left(x\right)=6x+2a$

$\Rightarrow f\text{'}\text{'}\text{'}\left(x\right)=6$

According to the question, 

$a=f\text{'}\left(1\right)=3+2a+b\Rightarrow a+b=-3 ...\left(i\right)$

$b=f\text{'}\text{'}\left(2\right)=12+2a\Rightarrow 2a-b=-12 ...\left(ii\right)$

$c=f\text{'}\text{'}\text{'}\left(3\right)\Rightarrow c=6$

Solving equations $\left(i\right) \& \left(ii\right)$, we get, $a= -5 \& b=2$

$\Rightarrow f\left(x\right)=x^3-5x^2+2x+6$

$\therefore f\left(2\right)=8-20+4+6=-2.$