\(\begin{array}{l}\overrightarrow {MA} + \overrightarrow {MB} = \overrightarrow 0 \Leftrightarrow \overrightarrow {MA} + \overrightarrow {MA} + \overrightarrow {AB} = \overrightarrow 0 \Leftrightarrow \overrightarrow {AM} = \frac{1}{2}\overrightarrow {AB} \\\overrightarrow {NA} + 2\overrightarrow {NC} = \overrightarrow 0 \Leftrightarrow \overrightarrow {NA} + 2\left( {\overrightarrow {NA} + \overrightarrow {AC} } \right) = \overrightarrow 0 \Leftrightarrow \overrightarrow {AN} = \frac{2}{3}\overrightarrow {AC} \\3\overrightarrow {PB} + 4\overrightarrow {PC} = \overrightarrow 0 \Leftrightarrow 3\overrightarrow {PB} + 4\left( {\overrightarrow {PB} + \overrightarrow {BC} } \right) = \overrightarrow 0 \Leftrightarrow \overrightarrow {BP} = \frac{4}{7}\overrightarrow {BC} \\\overrightarrow {AP} = \overrightarrow {AB} + \overrightarrow {BP} = \overrightarrow {AB} + \frac{4}{7}\overrightarrow {BC} = \overrightarrow {AB} + \frac{4}{7}\left( {\overrightarrow {AC} - \overrightarrow {AB} } \right) = \frac{3}{7}\overrightarrow {AB} + \frac{4}{7}\overrightarrow {AC} \\\overrightarrow {AG} = \frac{1}{3}\left( {\overrightarrow {AM} + \overrightarrow {AN} + \overrightarrow {AP} } \right) = \frac{1}{3}\left( {\frac{1}{2}\overrightarrow {AB} + \frac{2}{3}\overrightarrow {AC} + \frac{3}{7}\overrightarrow {AB} + \frac{4}{7}\overrightarrow {AC} } \right)\\ = \frac{1}{3}\left( {\frac{{13}}{{14}}\overrightarrow {AB} + \frac{{26}}{{21}}\overrightarrow {AC} } \right) = \frac{{13}}{{42}}\overrightarrow {AB} + \frac{{26}}{{63}}\overrightarrow {AC} \end{array}\)
Vì \(\frac{{\frac{3}{7}}}{{\frac{{13}}{{42}}}} = \frac{{\frac{4}{7}}}{{\frac{{26}}{{63}}}}\) nên \(\overrightarrow {AP} ,\overrightarrow {AG} \) cùng phương nên ba điểm \(A,G,P\) thẳng hàng.
