Let $A D$ and $B E$ are altitudes of the triangle.
$\therefore $ Equation of $A D$ is given by
$y-3 =\frac{-1}{\text { Slope of } B C}(x+2) $
$\Rightarrow y-3 =\frac{-1}{\left(\frac{0+1}{4-2}\right)}(x+2) $
$ y-3=-2(x+2) $
$ y-3=-2 x-4 $
$\Rightarrow 2 x+y+1=0$.......(i)
$\therefore $ Equation of $B E$ is given by
$ y+1=\frac{-1}{\text { Slope of } A C}(x-2)$
$\Rightarrow y+1=\frac{-1}{\left(\frac{0-3}{4+2}\right)}(x-2) $
$y+1=2(x-2)$
$y+1=2 \,x-4 $
$\Rightarrow y=2\, x-5$.......(ii)
Since, orthocentre is the intersecting point of altitudes.
$\therefore $ On solving Eqs. (i) and (ii), we get orthocentre $(1,-3)$
Also, centroid of $\triangle A B C=\left(\frac{-2+2+4}{3}, \frac{3-1+0}{3}\right)$
$=\left(\frac{4}{3}, \frac{2}{3}\right)$
$\therefore $ Equation of line joining $(0,-1)$ and $\left(\frac{4}{3}, \frac{2}{3}\right)$ is
$y+3=\frac{\frac{2}{3}+3}{\frac{4}{3}-1}(x-1) $
$ \Rightarrow y+3=11(x-1) $
$ \Rightarrow y+3=11 x-11 $
$ \Rightarrow 11\, x-y-14=0 $