Maximum possible number of points of intersection of 7 lines is ${ }^7{C}_2=21$ 

Maximum possible number of points of intersection of 5  circles in $2\left({ }^5{C}_2\right)=20$

$[\because$ two circles can intersect in two distinct points.]

Maximum possible number of points of intersection of 5 circles and 7 lines is 2(5) (7) = 70

Thus, maximum possible number of points of intersection is  21 + 20 + 70 = 111.