One zero after every 10 numbers, thus 20 + 10 extra (for 100’s series, 100, 101 to 109) + 1 extra for 200
Thus total number of zeros between 1 to 200 are 31.

If we try to make equations

x1 + x2 + x3 = 9

x4 + x5 + 5 = 12

x7 + x8 + x9 = 7

x1 + x4 + x7 = 6

x2 + x5 + x8 = 9

x3 + 5 + x9 = 13

x1 + x5 + x9 = 5

x7 + x5 + x3 = 11

7 variables and 7 equations, its little tough and cumbersome to solve these equations so lets try some hit and trial.

you can see that x1 + x5 + x9 is only 5, thus it has less number of possibilities for x1, x5 and x9 as these cant be -ve.

lets assume x1 and x9 both are 0, in that case x5 will be 5.

17
0 1 8 9
2 5 5 12
4 3 0 7
——
6 9 13 5

diagonal equation does not satisfy, we can see x7 + x5 + x3 exceeds by 6.
so lets try to reduce it by 6,

lets try by x1 = 1, x9 = 1
11
1 1 7 9
4 3 5 12
1 5 1 7
6 9 13 5

it satisfies all the conditions
so, x1 = 1 , x2 = 1, x3 = 7, x4 = 4, x5 = 3, x7 = 1, x8 =5 and x9 = 1