Đáp án:
$\begin{array}{l}
a)\overrightarrow {AB} = \left( {2;2} \right);\overrightarrow {AC} = \left( {1; - 3} \right)\\
\Rightarrow \overrightarrow {AB} .\overrightarrow {AC} = 2.1 + 2.\left( { - 3} \right) = - 4 \ne \overrightarrow 0 \\
\Rightarrow A,B,C\,\,ko\,thẳng\,hàng\\
\Rightarrow ABC\,là\,1\,tam\,giác\\
b)G = \left( {\frac{{1 + 3 + 2}}{3};\frac{{2 + 4 - 1}}{3}} \right) = \left( {2;\frac{5}{3}} \right)\\
H\left( {x;y} \right) \Rightarrow \overrightarrow {AH} = \left( {x - 1;y - 2} \right);\overrightarrow {BH} = \left( {x - 3;y - 4} \right)\\
co:AH \bot BC\\
\Rightarrow \overrightarrow {AH} .\overrightarrow {BC} = \overrightarrow 0 \\
\Rightarrow \left( {x - 1} \right).\left( { - 1} \right) + \left( {y - 2} \right).\left( { - 5} \right) = 0\\
\Rightarrow - x - 5y + 11 = 0\left( 1 \right)\\
BH \bot AC \Rightarrow \overrightarrow {BH} .\overrightarrow {AC} = \overrightarrow 0 \\
\Rightarrow \left( {x - 3} \right).1 + \left( {y - 4} \right).\left( { - 3} \right) = 0\\
\Rightarrow x - 3y + 9 = 0\left( 2 \right)\\
Từ\,\left( 1 \right);\left( 2 \right) \Rightarrow x = \frac{{ - 3}}{2};y = \frac{5}{2} \Rightarrow H\left( {\frac{{ - 3}}{2};\frac{5}{2}} \right)
\end{array}$
