1.
$\vec{AB}(-2-1; 3-2)=(-3;1)$
$\vec{AC}(2-1; -1-2)=(1;-3)$
Ta thấy $\dfrac{-3}{1}\ne \dfrac{1}{-3}$
$\Rightarrow \vec{AB}\ne k\vec{AC}$
Vậy ba điểm $A, B, C$ không thẳng hàng.
2.
$m\vec{AC}=(m;-3m)$
$\Rightarrow \vec{AB}+m\vec{AC}=(-3+m; 1-3m)=(m-3; -3m+1)$
$\Rightarrow |\vec{AB}+m\vec{AC}|=\sqrt{(m-3)^2+(-3m+1)^2}$
$=\sqrt{m^2-6m+9+9m^2-6m+1}$
$=\sqrt{10m^2-12m+10}=M$
Ta có $10m^2-12m+10=(\sqrt{10}m)^2-2\sqrt{10}m.\dfrac{6}{\sqrt{10}}+\dfrac{18}{5}+\dfrac{32}{5}$
$=(\sqrt{10}m-\dfrac{6}{\sqrt{10}})^2+\dfrac{32}{5}\ge \dfrac{32}{5}$
$\Rightarrow A\ge \dfrac{4\sqrt{10}}{5}$
$\min A=\dfrac{4\sqrt{10}}{5}\Leftrightarrow \sqrt{10}m-\dfrac{6}{\sqrt{10}}=0$
$\Leftrightarrow m=\dfrac{3}{5}$