$\begin{array}{l} 2\overrightarrow {MA} - 3\overrightarrow {MB} = 3\overrightarrow {BC} \\ \Leftrightarrow 2\overrightarrow {MA} = 3\overrightarrow {BC} + 3\overrightarrow {MB} \\ \Leftrightarrow 2\overrightarrow {MA} = 3\overrightarrow {BC} - 3\overrightarrow {BM} \\ \Leftrightarrow 2\overrightarrow {MA} = 3\overrightarrow {MC} \\ \Leftrightarrow \overrightarrow {MA} = \dfrac{3}{2}\overrightarrow {MC} \\ \Leftrightarrow \overrightarrow {MC} + \overrightarrow {CA} = \dfrac{3}{2}\overrightarrow {MC} \\ \Leftrightarrow \overrightarrow {CA} = \dfrac{3}{2}\overrightarrow {MC} - \overrightarrow {MC} = \dfrac{1}{2}\overrightarrow {MC} \end{array}$
Vậy $C$ nằm giữa $A,M$ thỏa mãn $MC=2CA$
