$$\eqalign{
& A\left( {1;1} \right);\,\,B\left( { - 2;7} \right);\,\,C\left( {3; - 3} \right);\,\,D\left( {3;2} \right) \cr
& a)\,\,\overrightarrow {AB} = \left( { - 3;6} \right);\,\,\overrightarrow {AC} = \left( {2; - 4} \right) \cr
& \Rightarrow \overrightarrow {AB} = - {3 \over 2}\overrightarrow {AC} \cr
& \Rightarrow A,\,\,B,\,\,C\,\,thang\,\,hang. \cr
& b)\,\,\overrightarrow {AD} = \left( {2;1} \right) \cr
& Gia\,\,su\,\,\exists k \in R,\,\,k \ne 0:\,\,\overrightarrow {AC} = k\overrightarrow {AD} \cr
& \Rightarrow \left\{ \matrix{
2 = k.2 \hfill \cr
- 4 = k.1 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
k = 1 \hfill \cr
k = - {1 \over 4} \hfill \cr} \right.\,\,\left( {Vo\,\,nghiem} \right) \cr
& \Rightarrow Khong\,\,ton\,\,tai\,\,hang\,\,so\,\,k\,\,de:\,\overrightarrow {AC} = k\overrightarrow {AD} \cr
& \Rightarrow A,\,\,C,\,\,\,D\,\,khong\,\,thang\,\,hang. \cr
& c)\,\,AC = \sqrt {{2^2} + {{\left( { - 4} \right)}^2}} = 2\sqrt 5 \cr
& \,\,\,\,\,\,AD = \sqrt {{2^2} + {1^2}} = \sqrt 5 \cr
& \,\,\,\,\,\,CD = \sqrt {{0^2} + {5^2}} = 5 \cr
& Ta\,\,co:\,\,A{C^2} + A{D^2} = 20 + 5 = 25 = C{D^2} \cr
& \Rightarrow \Delta ACD\,\,vuong\,\,\,tai\,\,A\,\,\left( {Dinh\,\,li\,\,Pytago\,\,dao} \right) \cr
& {S_{ADC}} = {1 \over 2}.AC.AD = {1 \over 2}.2\sqrt 5 .\sqrt 5 = 5 \cr} $$
