The number of seven digit integers with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only is:

We need to find seven-digit integers using only digits 1, 2, and 3 whose digits sum to 10. Let $$n_1$$, $$n_2$$, and $$n_3$$ denote the number of 1s, 2s, and 3s respectively. Then $$n_1 + n_2 + n_3 = 7$$ and $$n_1 + 2n_2 + 3n_3 = 10$$.

Subtracting the first equation from the second gives $$n_2 + 2n_3 = 3$$. The non-negative integer solutions are: $$(n_3 = 0, n_2 = 3, n_1 = 4)$$ and $$(n_3 = 1, n_2 = 1, n_1 = 5)$$.

For the first case, the number of arrangements of 4 ones and 3 twos in 7 positions is $$\frac{7!}{4! \cdot 3!} = 35$$.

For the second case, the number of arrangements of 5 ones, 1 two, and 1 three in 7 positions is $$\frac{7!}{5! \cdot 1! \cdot 1!} = 42$$.

The total number of such seven-digit integers is $$35 + 42 = 77$$.