Sorry.May I correct a part of the ending ? :

.....

\(\Rightarrow\left(m-5\right)^2-2^2=0\Rightarrow\left(m-7\right)\left(m-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}m=7\\m=3\end{matrix}\right.\)

Since the equation has 2 distinct roots,we have :

\(\left[-2\left(m+1\right)\right]^2-4.3.\left(3m-5\right)>0\)

\(\Leftrightarrow4\left(m^2+2m+1\right)-12\left(3m-5\right)>0\)

\(\Leftrightarrow4m^2+8m+4-36m+60>0\)

\(\Leftrightarrow\left(2m\right)^2-2.2m.7+7^2+15>0\)

\(\Leftrightarrow\left(2m-7\right)^2+15>0\) (always right with any m)

So,there's no condition.Let x₁ and x₂ be the roots (x₁ = 3x₂)

By applying the Viete's formula,we have :

\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{2\left(m+1\right)}{3}\\x_1x_2=\dfrac{3m-5}{3}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4x_2=\dfrac{2\left(m+1\right)}{3}\\3x_2^2=\dfrac{3m-5}{3}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x_2=\dfrac{m+1}{6}\\x_2^2=\dfrac{3m-5}{9}\end{matrix}\right.\)\(\Rightarrow\left(\dfrac{m+1}{6}\right)^2=\dfrac{3m-5}{9}\)

\(\Rightarrow9\left(m+1\right)^2-36\left(3m-5\right)=0\)\(\Rightarrow\left(m+1\right)^2-4\left(3m-5\right)=0\)

\(\Rightarrow m^2+2m+1-12m+20=0\)\(\Rightarrow m^2-2m.5+5^2-4=0\)

\(\Rightarrow\left(m+5\right)^2-2^2=0\)\(\Rightarrow\left(m+3\right)\left(m+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}m=-3\\m=-7\end{matrix}\right.\)