So we can choose the first two numbers how we like. The third has to have a definite residue class mod $3$ to make the total divisible by $3$.
If the third residue class is distinct from the residue class the first two numbers, the first two have to be from different residue classes. The number of ways of choosing one from each residue class is $\binom {300}1 \times \binom {200}1 \times\binom {100}1$, but there are six different orders in which the same three numbers can be selected.
If the final residue class is the same as one of the previous ones, they all have to be the same. We choose one of the three residue classes, and there are then $\binom {100}3$ ways of choosing a triple.
So the overall number of ways is $$\frac 16\times\binom {300}1 \times \binom {200}1 \times\binom {100}1+3\times\binom {100}3$$
And this is equal to $$\binom {100}1 \times \binom {100}1 \times\binom {100}1+3\times\binom {100}3$$
I think the analogy with the permutations of letters is making this problem more complicated than it needs to be.
Using the restriction that the number has at least one seven, you can first find the numbers that have exactly one $7$, then the numbers that have two $7$s, and then the number that has three $7$s and then add the results.
To find the number of numbers, think of choosing a digit for each spot: _ _ _
For one seven, you can fix a $7$ in a spot, say the first one, so the number looks like 7_ _ and note that for the other spots you can have any of the other 9 digits ($0,1,2,3,4,5,6,8,$ or $9$), so there are $9\cdot 9=81$ such numbers.
For numbers of the forms _ 7 _ and _ _ 7 the count is different because the first digit cannot be $0$, so there are $8\cdot 9=72$ possibilities for each.
Thus, in total there are $72+72+81=225$ three-digit positive integers with one seven as a digit.
Two sevens: For the form _77 there are 8 possibilities because the first spot cannot be 0, and for each of the forms 7_7 and _ _7 there are 9 possibilities, so in total there are $8+9+9=26$ three-digit positive integers with one seven as a digit.
Three sevens: There is only one, $777$.
So in total there are $225+26+1=252$ three-digit integers with a seven as a digit.
Best Answer
Note that in order that the sum of $3$ number will be divided by $3$, you have that their sum modulo $3$ will be $0$, therefore you could have the next possibilities:
it's not hard to calculate each one of those cases, and since they are disjoint the sum of them to get the wanted result.