G1.1 Given that $(a+b+c)^{2}=3\left(a^{2}+b^{2}+c^{2}\right)$ and $a+b+c=12$, find the value of $a$.
$\begin{array}{l}\text { Sub. (2) into }(1), 12^{2}=3\left(a^{2}+b^{2}+c^{2}\right) \\ \Rightarrow a^{2}+b^{2}+c^{2}=48 \ldots \ldots(3) \\ (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a) \\ \Rightarrow 12^{2}=48+2(a b+b c+c a) \\ \Rightarrow a b+b c+c a=48 \\ 2\left[a^{2}+b^{2}+c^{2}-(a b+b c+c a)\right]=(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \\ 2[48-48]=0=(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \\ \Rightarrow a=b=c \\ a+b+c=3 a=12 \\ \Rightarrow a=4\end{array}$