\(\overrightarrow{CB}=\left(7;3\right)\) ; \(\overrightarrow{MA}=\left(a+1;a-3\right)\)
d qua M \(\Leftrightarrow AM\perp BC\)
\(\Leftrightarrow\overrightarrow{CB}.\overrightarrow{MA}=0\)
\(\Leftrightarrow7\left(a+1\right)+3\left(a-3\right)=0\)
\(\Rightarrow a=\frac{1}{5}\)