Đáp án:
$\begin{array}{l}
C\left( {1; - 1;1} \right)\\
3)M\left( {x;y;z} \right)\\
\Rightarrow \overrightarrow {AM} = \left( {x - 1;y;z - 1} \right);\overrightarrow {MD} = \left( {4 - x;5 - y; - 5 - z} \right)\\
Do:\overrightarrow {AM} = \overrightarrow {MD} \\
\Rightarrow \left\{ \begin{array}{l}
x - 1 = 4 - x\\
y = 5 - y\\
z - 1 = - 5 - z
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = \frac{5}{2}\\
y = \frac{5}{2}\\
z = - 2
\end{array} \right. \Rightarrow M\left( {\frac{5}{2};\frac{5}{2}; - 2} \right)\\
N\left( {x;y} \right)\\
\Rightarrow \overrightarrow {AN} = \left( {x - 1;y;z - 1} \right);\overrightarrow {ND} = \left( {4 - x;5 - y; - 5 - z} \right)\\
Do:\overrightarrow {AN} = 2\overrightarrow {ND} \\
\Rightarrow \left\{ \begin{array}{l}
x - 1 = 2\left( {4 - x} \right)\\
y = 2\left( {5 - y} \right)\\
z - 1 = 2\left( { - 5 - z} \right)
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = \frac{7}{3}\\
y = \frac{7}{3}\\
z = - 3
\end{array} \right. \Rightarrow N\left( {\frac{7}{3};\frac{7}{3}; - 3} \right)\\
6)\\
\overrightarrow {BA} = \left( {1;1;1} \right);\overrightarrow {BC} = \left( { - 1; - 2; - 1} \right);\overrightarrow {CD} = \left( {3;5; - 6} \right)\\
\Rightarrow cos\widehat {ABC} = \frac{{\overrightarrow {BA} .\overrightarrow {BC} }}{{BA.BC}} = \frac{{ - 1 - 2 - 1}}{{\sqrt 3 .\sqrt {1 + 4 + 1} }} = \frac{{ - 2\sqrt 2 }}{3}\\
\Rightarrow \widehat {ABC} = {160^0}\\
\Rightarrow \widehat {BCD} = {70^0}\\
7)\\
I\left( {x;y;z} \right) \Rightarrow \overrightarrow {ID} = \left( {4 - x;5 - y; - 5 - z} \right)\\
Do:3\overrightarrow {AB} - \overrightarrow {ID} = 2\overrightarrow {CB} \\
\Rightarrow \overrightarrow {ID} = 3\overrightarrow {AB} - 2\overrightarrow {CB} = 3\overrightarrow {AB} + 2\overrightarrow {BC} = \left( { - 5; - 7; - 5} \right)\\
\Rightarrow \left\{ \begin{array}{l}
4 - x = - 5\\
5 - y = - 7\\
- 5 - z = - 5
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = 9\\
y = 12\\
z = 0
\end{array} \right.\\
\Rightarrow I\left( {9;12;0} \right)
\end{array}$
