Find`vec a . vec b`, When`vec a = hat i + hat j - 2 hat k` and `vec b = 3 hat i +2 hat j + hat k`

Explanation

The elements of A are all multiples of 5. Sum of every pair of elements of A is divisible by 5. Therefore, we have to find the probability that B has two distinct elements whose sum is divisible byy 3.
Let `A_(0)` be the set of elements of A of the form 3k, i.e., {0, 15, 30, ..., 195}, `A_(1)` be the set of elements of A of the form 3k + 1, i.e., {10, 25, ..., 190}, `A_(2)` be the set of elements of A of the form 3k + 2, i.e., {5, 20, 35, ..., 185}. Then `n(A_(0))` = 14, `n(A_(1)) = n(A_(2)) = 13`.
If B has at least two elements from `A_(0)`, then we are done.
If B contains at most one element of `A_(0)`, then it must have at least one element from each of `A_(1)` and `A_(2)` for which the sum of these two elements will be divisible by 3. So, the required probability is 1.