Lời giải

a) Ta có \[\overrightarrow {AB} = \left( {4;\,\,3} \right);\,\,\overrightarrow {AC} = \left( {6;\,\,1} \right)\]

Suy ra \[2\overrightarrow {AC} = \left( {12;\,\,2} \right)\]

Do đó \[\overrightarrow u = \left( {12 - 4;\,\,2 - 3} \right) = \left( {8;\,\, - 1} \right)\].

b) Ta có \[M\left( {x;\,\,y} \right) \Rightarrow \left\{ \begin{array}{l}\overrightarrow {MA} = \left( { - \,4 - x;\,\, - y} \right)\\\overrightarrow {MB} = \left( { - \,x;\,\,3 - y} \right)\\\overrightarrow {MC} = \left( {2 - x;\,\,1 - y} \right)\end{array} \right.\]

Khi đó \[\overrightarrow {MA} + 2\overrightarrow {MB} + 3\overrightarrow {MC} = \left( {2 - 6x;\,\,9 - 6y} \right)\]

\[ \Rightarrow \left\{ \begin{array}{l}2 - 6x = 0\\9 - 6y = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = \frac{1}{3}\\y = \frac{3}{2}\end{array} \right.\].

Do đó \[M\left( {\frac{1}{3};\,\,\frac{3}{2}} \right)\].