Let f

Let $f(x)=x^{3}+a x^{2}+b x+c$

$f^{\prime}(x)=3 x^{2}+2 a x+b \Rightarrow f^{\prime}(1)=3+2 a+b$

$f^{\prime \prime}(x)=6 x+2 a \Rightarrow f^{\prime \prime}(2)=12+2 a$

$f^{\prime \prime \prime}(x)=6 \Rightarrow f^{\prime \prime \prime}(3)=6$

$\because f(x)=x^{3}+f^{\prime}(1) x^{2}+f^{\prime \prime}(2) x+f^{\prime \prime \prime}(3)$

$\therefore f^{\prime}(1)=a \Rightarrow 3+2 a+b=a \Rightarrow a+b=-3$ $\ldots(1)$

alsof" $(2)=b \Rightarrow 12+2 a=b \Rightarrow 2 a-b=-12$ $\ldots(2)$

and $f^{\prime \prime \prime}(3)=c \Rightarrow \mathrm{c}=6$

Add (1) and (2)

$3 a=-15 \Rightarrow a=-5 \Rightarrow b=2$

$\Rightarrow f(x)=x^{3}-5 x^{2}+2 x+6$

$\Rightarrow f(2)=8-20+4+6=-2$