\(\overrightarrow{AB}=\left(3;2\right)\Rightarrow\overrightarrow{n_p}=\left(-2;3\right)\)
\(\Rightarrow\left(AB\right):-2\left(x+3\right)+3y=0\Leftrightarrow2x-3y+6=0\)
\(\overrightarrow{BC}=\left(3;-1\right)\Rightarrow\overrightarrow{n_p}=\left(1;3\right)\)
\(\Rightarrow\left(BC\right):x+3\left(y-2\right)=0\Leftrightarrow x+3y-6=0\)
\(\overrightarrow{CD}=\left(0;-3\right)\Rightarrow\overrightarrow{n_p}=\left(3;0\right)\)
\(\Rightarrow\left(CD\right):3\left(x-3\right)=0\Leftrightarrow x-3=0\)
\(\overrightarrow{DA}=\left(-6;2\right)\Rightarrow\overrightarrow{n_p}=\left(-2;-6\right)=2\left(-1;-3\right)\)
\(\Rightarrow\left(DA\right):-\left(x+3\right)-3y=0\Leftrightarrow x+3y+3=0\)
Để \(M\left(m;m-1\right)\) nằm trong tứ giác \(ABCD\) 4 cạnh khi và chỉ khi
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y>0\\x+3y-6< 0\\x-3< 0\\x+3y+3>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2m-3\left(m-1\right)>0\\m+3\left(m-1\right)-6< 0\\m-3< 0\\m+3\left(m-1\right)+3>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m< 3\\m< \dfrac{9}{4}\\m< 3\\m>0\end{matrix}\right.\) \(\Leftrightarrow0< m< \dfrac{9}{4}\)
Vậy \(0< m< \dfrac{9}{4}\) thỏa mãn yêu cầu đề bài