G2.1 Given that the equation $a x(x+1)+b x(x+2)+c(x+1)(x+2)=0$ has roots 1 and 2 . If $a+b+c=2$, find the value of $a$.
Expand and rearrange the terms in descending orders of $x$ :
$$
\begin{array}{l}
(a+b+c) x^{2}+(a+2 b+3 c) x+2 c=0 \\
2 x^{2}+(a+b+c+b+2 c) x+2 c=0 \\
2 x^{2}+(b+2 c+2) x+2 c=0
\end{array}
$$

It is identical to $2(x-1)(x-2)=0$
$$
\therefore b+2 c+2=-6 ; 2 c=4
$$

Solving these equations give $c=2, b=-12, a=12$