No. of ways of selecting 'r' consecutive things from 'n' consecutive things = n - r + 1

So, you can select 3 consecutive numbers in 100 - 3 + 1 = 98 ways.

Now, out of 98 selections: you can get either of the two combinations:

i) even - odd - even: If out of numbers chosen two are even and one is odd, then certainly the product of these numbers will be divisible by 24. No. of ways we can select even-odd-even is $\frac{98}{2}$ = 49

ii) odd - even - odd: The product of two odd numbers and one even number will be divisible by 12 if the even number is  a multiple of 8. We have 12 even multiples of 8 within 100. Hence we can have 12 combinations of odd- even - odd.

So, total number of favourable cases: 49 + 12 = 61

So, probability = $\frac{61}{98}$. Hence option (c)