A line $y=mx+1$  intersects the circle ${\left(x-3\right)}^2+{\left(y+2\right)}^2=25$  at the points P and Q. If the midpoint of the line segment PQ has $x-$-coordinate $-\frac{3}{5}$  then which one of the following options is correct ?

$\text{The midpoint of the chord PQ\hspace{0.17em}is R }\left(-\frac{3}{5},\frac{-3m+5}{5}\right)$

$$\begin{gathered} \text{The centre of the circle is C}\left(3,-2\right) \\ CR\hspace{0.17em}is perpendicular to PQ \end{gathered}$$

$\text{The product of the slopes of perpendicular lines is -1}$

$\begin{array}{rcl}m\times \left(\frac{\frac{-3m+5}{5}+2}{-\frac{3}{5}-3}\right)& =& -1\\ m\times \frac{-3m+15}{-18}& =& -1\\ 3m^2-15m+18& =& 0\end{array}$

$Solving the equation$

$\begin{array}{rcl}{m}^2-5m+6& =& 0\\ m^2-3m-2m+6& =& 0\\ \left(m-2\right)\left(m-3\right)& =& 0\\ m& =& 2,3\end{array}$

$\text{Threfore, from the options 2≤m≤4}$