Equation of circle is
$x^2+y^2-6 x+4 y-12=0$
$\therefore \quad 2 x+2 y y^{\prime}-6+4 y^{\prime}=0$
$\Rightarrow \quad y^{\prime}=\frac{6-2 x}{2 y+4}$
$\begin{gathered}\therefore \text { Slope of tangent at }(-1,1)=\left.\frac{d v}{d x}\right|_{(-1,1)} \\ =\frac{6-2(-1)}{2(1)+4}=\frac{8}{6}=\frac{4}{3}\end{gathered}$
$\therefore$ Equation of tangent is
$\begin{aligned} & & y-1=\frac{4}{3}(x+1) \\ \Rightarrow & & 3 y-3=4 x+4 \\ \Rightarrow & & 4 x-3 y+7=0\end{aligned}$
Let the equation of chord is
$4 x-3 y+k=0$
$\begin{array}{ll}\therefore & \frac{k-7}{\sqrt{16+9}}= \pm 1 \Rightarrow k-7= \pm 5 \\ \Rightarrow & k=7 \pm 5 \Rightarrow k=12,2\end{array}$
$\therefore$ Equation of chord can be
$4 x-3 y+12=0$ or $4 x-3 y+2=0$
Now, Distance of both above lines from centre of circle are
$d_1=\left|\frac{12+6+12}{5}\right|=6$ and $\quad d_2=\left|\frac{12+6+2}{5}\right|=4$
and radius of circle, $r=\sqrt{9+4+12}=5$
$\therefore d_2 < r$
$\therefore$ Equation of chord is $4 x-3 y+2=0$

Now, slope of line $4 x-3 y+2=0$ is $\frac{4}{3}$
$\therefore$ Slope of line $C P=\frac{-3}{4}$
$\therefore$ Equation of CP is
$\begin{aligned} y+2 & =\frac{-3}{4}(x-3) \\ \Rightarrow \quad 4 y+8 & =-3 x+9 \Rightarrow 3 x+4 y=1\end{aligned}$
On solving $4 x-3 y+2=0$ and $3 x+4 y=1$, we get $P\left(-\frac{1}{5}, \frac{2}{5}\right)$
$\therefore$ Mid-point of chord is $\left(-\frac{1}{5}, \frac{2}{5}\right)$