$$
\begin{array}{ll}
& \mathrm{f}(x)=x^3+x^2 \mathrm{f}^{\prime}(1)+x \mathrm{f}^{\prime \prime}(2)+6 \\
\therefore \quad & \mathrm{f}^{\prime}(x)=3 x^2+2 x \mathrm{f}^{\prime}(1)+\mathrm{f}^{\prime \prime}(2) \\
\therefore \quad & \mathrm{f}^{\prime \prime}(x)=6 x+2 \mathrm{f}^{\prime}(1)
\end{array}
$$
Substituting $x=1$ in (i), we get
$$
\begin{aligned}
& \mathrm{f}^{\prime}(1)=3(1)^2+2(1) \mathrm{f}^{\prime}(1)+\mathrm{f}^{\prime \prime}(2) \\
& \Rightarrow \mathrm{f}^{\prime}(1)+\mathrm{f}^{\prime \prime}(2)=-3
\end{aligned}
$$
Substituting $x=2$ in (ii), we get
$$
\begin{aligned}
& \mathrm{f}^{\prime \prime}(2)=6(2)+2 \mathrm{f}^{\prime}(1) \\
& \Rightarrow \mathrm{f}^{\prime \prime}(2)=12+2 \mathrm{f}^{\prime}(1)
\end{aligned}
$$
From (iii) and (iv), we get
$$
\begin{aligned}
& \mathrm{f}^{\prime}(1)+12+2 \mathrm{f}^{\prime}(1)=-3 \\
& \Rightarrow 3 \mathrm{f}^{\prime}(1)=-15 \\
& \Rightarrow \mathrm{f}^{\prime}(1)=-5
\end{aligned}
$$
From (iii), $-5+\mathrm{f}^{\prime \prime}(2)=-3$
$\Rightarrow \mathrm{f}^{\prime \prime}(2)=2$
$$
\begin{aligned}
\therefore \quad \mathrm{f}(2) & =2^3+2^2(-5)+2(2)+6 \\
& =8-20+4+6=-2
\end{aligned}
$$