Đáp án:
$\begin{array}{l}
a)\left\{ \begin{array}{l}
{x_G} = \dfrac{{{x_A} + {x_B} + {x_C}}}{3} = \dfrac{{1 + 3 + 6}}{3} = \dfrac{{10}}{3}\\
{y_G} = \dfrac{{{y_A} + {y_B} + {y_C}}}{3} = \dfrac{{4 - 1 + 2}}{3} = \dfrac{5}{3}
\end{array} \right.\\
\Rightarrow G\left( {\dfrac{{10}}{3};\dfrac{5}{3}} \right)
\end{array}$
b) Gọi trung điểm của BC và CA là M và N
$\begin{array}{l}
\Rightarrow \left\{ \begin{array}{l}
{x_M} = \dfrac{{{x_B} + {x_C}}}{2} = \dfrac{{3 + 6}}{2} = \dfrac{9}{2}\\
{y_M} = \dfrac{{{y_B} + {y_C}}}{2} = \dfrac{{ - 1 + 2}}{2} = \dfrac{1}{2}
\end{array} \right.\\
\Rightarrow M\left( {\dfrac{9}{2};\dfrac{1}{2}} \right)\\
\left\{ \begin{array}{l}
{x_N} = \dfrac{{{x_A} + {x_C}}}{2} = \dfrac{{1 + 6}}{2} = \dfrac{7}{2}\\
{y_N} = \dfrac{{{y_A} + {y_C}}}{2} = \dfrac{{4 + 2}}{2} = 3
\end{array} \right.\\
\Rightarrow N\left( {\dfrac{7}{2};3} \right)\\
Gọi\,MN:y = a.x + b\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{1}{2} = \dfrac{9}{2}.a + b\\
3 = \dfrac{7}{2}a + b
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = - \dfrac{5}{2}\\
b = \dfrac{{47}}{4}
\end{array} \right.\\
\Rightarrow MN:y = - \dfrac{5}{2}x + \dfrac{{47}}{4}\\
c)M \in \left( d \right):3x - y + 1 = 0\\
\Rightarrow 3{x_M} - {y_M} + 1 = 0\\
\Rightarrow M\left( {{x_M};3{x_M} + 1} \right)\\
AM = 5\\
\Rightarrow A{M^2} = 25\\
\Rightarrow {\left( {{x_M} - 1} \right)^2} + {\left( {3{x_M} + 1 - 4} \right)^2} = 25\\
\Rightarrow 10{\left( {{x_M} - 1} \right)^2} = 25\\
\Rightarrow {\left( {{x_M} - 1} \right)^2} = \dfrac{5}{2}\\
\Rightarrow \left[ \begin{array}{l}
{x_M} = \dfrac{{\sqrt {10} }}{2} + 1 \Rightarrow {y_M} = \dfrac{{3\sqrt {10} }}{2} + 4\\
{x_M} = - \dfrac{{\sqrt {10} }}{2} + 1 \Rightarrow {y_M} = \dfrac{{ - 3\sqrt {10} }}{2} + 4
\end{array} \right.\\
\Rightarrow M\left( {\dfrac{{ \pm \sqrt {10} }}{2} + 1;\dfrac{{ \pm 3\sqrt {10} }}{2} + 4} \right)
\end{array}$
