Given,

The line $2x-3y+4=0$ divides the line segment joining the points $A\left(-2,3\right)$ and $B\left(3,-2\right)$ in the ratio $m:n$,

Now let the point be $P$ which divide in ratio $m:n$,

So by section formula we get $P\equiv \left(\frac{3m-2n}{m+n},\frac{3n-2m}{m+n}\right)$

Now, given point $P$ lie on the line $2x-3y+4=0$ 

So $2\times \frac{3m-2n}{m+n}-3\times \frac{3n-2m}{m+n}+4=0$

$\Rightarrow \frac{m}{n}=\frac{9}{16}$

Then the point which divides $AB$ in the ratio $\frac{-4 \mathrm{m}}{3\mathrm{n}}=\frac{-4}{3}\times \frac{9}{16}=\frac{-3}{4}$ by section formula will be,

$\left(\frac{-3\times 3-8}{1},\frac{6+12}{1}\right)\equiv \left(-17,18\right)$