Given,

The line $x-y+1=0$ cuts the lines $2x+2y+3=0$ and $3x+3y+2=0$ at the points $A$ and $B$ respectively,

Now we can see that $2x+2y+3=0$ slope $=1$ and $3x+3y+2=0$ slope $=1$ are parallel lines as their slopes are equal,

Now line $x-y+1=0$ has slope $=-1$,

So  $x-y+1=0$ is perpendicular to both $2x+2y+3=0 \& 3x+3y+2=0$,

So the distance between $AB$ is same as distance between parallel lines,

So $AB=\left|\frac{{\displaystyle \frac{2}{3}}-{\displaystyle \frac{3}{2}}}{\sqrt{1^2+1^2}}\right|=\frac{5}{6\sqrt{5}}$