Đáp án:

$\begin{array}{l}
a)\overrightarrow {AB}  = \left( { - 3;1} \right);\overrightarrow {AC}  = \left( {1; - 1} \right);\overrightarrow {BC}  = \left( {4; - 2} \right)\\
\left\{ \begin{array}{l}
{x_G} = \frac{{{x_A} + {x_B} + {x_C}}}{3} = \frac{{1 - 2 + 2}}{3} = \frac{1}{3}\\
{y_G} = \frac{{{y_A} + {y_B} + {y_C}}}{3} = \frac{{2 + 3 + 1}}{3} = 2
\end{array} \right. \Rightarrow G\left( {\frac{1}{3};2} \right)\\
H\left( {x;y} \right)\\
 \Rightarrow \overrightarrow {BH}  = \left( {x + 2;y - 3} \right);\overrightarrow {CH}  = \left( {x - 2;y - 1} \right)\\
 \Rightarrow \left\{ \begin{array}{l}
BH \bot AC\\
CH \bot AB
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
\overrightarrow {BH} .\overrightarrow {AC}  = 0\\
\overrightarrow {CH} .\overrightarrow {AB}  = 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x + 2 - \left( {y - 3} \right) = 0\\
 - 3\left( {x - 2} \right) + \left( {y - 1} \right) = 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x = 5\\
y = 10
\end{array} \right. \Rightarrow H\left( {5;10} \right)\\
I\left( {x;y} \right)\\
 \Rightarrow IA = IB = IC\\
 \Rightarrow \left\{ \begin{array}{l}
I{A^2} = I{B^2}\\
I{A^2} = I{C^2}
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
{\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} = {\left( {x + 2} \right)^2} + {\left( {y - 3} \right)^2}\\
{\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} = {\left( {x - 2} \right)^2} + {\left( {y - 1} \right)^2}
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x =  - 2\\
y =  - 2
\end{array} \right. \Rightarrow I\left( { - 2; - 2} \right)\\
b)D\left( {x;y} \right)\\
 \Rightarrow \overrightarrow {AB}  = \overrightarrow {DC} \\
 \Rightarrow \left\{ \begin{array}{l}
2 - x =  - 3\\
1 - y = 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = 5\\
y = 0
\end{array} \right. \Rightarrow D\left( {5;0} \right)\\
c)\,H\left( {x;y} \right) \Rightarrow \overrightarrow {AH}  = \left( {x - 1;y - 2} \right)\\
Gọi\,pt\,dt\,BC:y = ax + b\\
B,C \in BC \Rightarrow \left\{ \begin{array}{l}
3 = a.\left( { - 2} \right) + b\\
1 = 2a + b
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a =  - \frac{1}{2}\\
b = 2
\end{array} \right.\\
 \Rightarrow y =  - \frac{1}{2}x + 2\\
H \in BC \Rightarrow {y_H} =  - \frac{1}{2}{x_H} + 2\\
Mà:AH \bot BC\\
 \Rightarrow 4\left( {x - 1} \right) - 2\left( {y - 2} \right) = 0\\
 \Rightarrow 4x - 4 - 2.\left( { - \frac{1}{2}x + 2 - 2} \right) = 0\\
 \Rightarrow 4x - 4 + x = 0\\
 \Rightarrow x = \frac{4}{5} \Rightarrow y = \frac{8}{5}\\
 \Rightarrow H\left( {\frac{4}{5};\frac{8}{5}} \right)
\end{array}$