$\text{ The sum of all integral value(s) of ' }r\text{ ' for which the circles }$

$x^2+y^2-10x+16y+89-r^2=0\text{ and }{x}^2+y^2+6x-14y+42=0\text{ intersect in two real }$$\text{distinct points is}$

$\begin{array}{l}\text{ We have }(x-5{)}^2+(y+8{)}^2=25+64+r^2-89\\ \text{ And }(x+3{)}^2+(y-7{)}^2=49+9-42=16\end{array}$

$\begin{array}{l}\Rightarrow (x-5{)}^2+(y+8{)}^2=r^2\text{ and }(x+3{)}^2+(y-7{)}^2=(4{)}^2\\ \Rightarrow (5,-8)(-3,7)\\ \Rightarrow \sqrt{65+225}=\sqrt{289}=17\end{array}$

$=\text{ distance between their centres }$

$\begin{array}{l}\text{ Now, }|r-4|<17<r+4\Rightarrow r+4>17\Rightarrow r>13\\ \text{ And }-17<r-4<17\Rightarrow -13<r<21\\ \text{ Hence }13<r<21\end{array}$

${\text{ Possible values of }}^{\mathrm{\prime }}{r}^{\mathrm{\prime }}\text{ can be }14,15,16,17,18,19,20$

Hence required sum = 119.