$(a,b,c)=\fbox{(3,223,9)}$ Summing both of the equations we get $\implies a+b+bc+ac=2260 $ $\implies (a+b)(1+c)=2260$ So $1+c|2260$ . Possible values of $1+c$ is $2,4,5,10,20,113,226,453,565,1130,2260$ Again Subtracting both of the equations we get $a-b-ac+ab=1760$ $\implies (b-a)(c-1)=1760$ So $c-1|1760$. Now possible values of $c-1$ is $1,2,4,5,8,10,11,16,20,22,32,40,44,55,80,88,110,160,176,220,352,440,880$ Considering these possible values of $c+1$ and $c-1$ we see that only $c=9$ works. For $c=9$ we get $a+b=226$ and $b-a=220$. $b=223$ and $a=3$ satisfy these equations. Thus $(a,b,c)=(3,223,9)$