The distance between $\left(1,1\right)$ and the point of intersection of two lines $2x+3y+c=0$ and  $3x+2y+c=0$ is less than $2\sqrt{2}$then 

To get the point of intersection of two lines, eliminate  $x$

$\begin{array}{l}3\left(2x+3y+c\right)-2\left(3x+2y+c\right)=0\\ 6x+9y+3c-6x-4y-2c=0\\ 5y+c=0\\ y=-\frac{c}{5}\end{array}$

$\text{ Substitute }y=-\frac{c}{5}\text{ in the equation }2x+3y+c=0$

$\begin{array}{l}2x+3\left(-\frac{c}{5}\right)+c=0\\ 10x-3c+5c=0\\ 10x+2c=0\\ x=-\frac{c}{5}\end{array}$

$\text{ Hence, the point of intersection is }\left(-\frac{c}{5},-\frac{c}{5}\right)$

$\text{ Given that the distance between }(1,1)\text{ and }\left(-\frac{c}{5},-\frac{c}{5}\right)\text{ is }\sqrt{{\left(1+\frac{c}{5}\right)}^2+{\left(1+\frac{c}{5}\right)}^2}<2\sqrt{2}$

It implies that 

$\sqrt{2{\left(1+\frac{c}{5}\right)}^2}<2\sqrt{2}$
  Squaring and then simplify 
  $\begin{array}{c}2{\left(1+\frac{c}{5}\right)}^2<8\\ {\left(1+\frac{c}{5}\right)}^2<4\end{array}$
Apply square root on both sides 

$\begin{array}{c}-2<1+\frac{c}{5}<2\\ -3<\frac{c}{5}<1\\ -15<c<5\end{array}$

$\text{ Therefore, }c\in (-15,5)$