Step-by-step Solution:

1. Understand the constraints and set up equations. We are looking for the number of 7-digit integers formed using only the digits 1, 2, and 3. The sum of the digits must be 10.

   Let:

   • x be the number of times the digit 1 appears.
   • y be the number of times the digit 2 appears.
   • z be the number of times the digit 3 appears.

   Based on the problem statement, we can form a system of two linear equations with three variables:

   • The total number of digits is 7: $x + y + z = 7 \quad \text{(Equation 1)}$
   • The sum of the digits is 10: $1 \cdot x + 2 \cdot y + 3 \cdot z = 10 \quad \text{(Equation 2)}$

   We need to find the number of non-negative integer solutions (x, y, z) for this system.

2. Solve the system of equations. To solve for x, y, and z, we can eliminate one variable. Let's subtract Equation 1 from Equation 2: $(x + 2y + 3z) - (x + y + z) = 10 - 7$ $y + 2z = 3$

   Since y and z must be non-negative integers (as they represent the count of digits), we can find the possible values by testing values for z starting from 0.

   • Case 1: If z = 0 Substituting z=0 into y + 2z = 3 gives y = 3. Now, substitute y=3 and z=0 into Equation 1: x + 3 + 0 = 7, which gives x = 4. So, one possible solution is (x, y, z) = (4, 3, 0). This means the digits are four 1s, three 2s, and zero 3s. The set of digits is {1, 1, 1, 1, 2, 2, 2}.

   • Case 2: If z = 1 Substituting z=1 into y + 2z = 3 gives y + 2(1) = 3, so y = 1. Now, substitute y=1 and z=1 into Equation 1: x + 1 + 1 = 7, which gives x = 5. So, another possible solution is (x, y, z) = (5, 1, 1). This means the digits are five 1s, one 2, and one 3. The set of digits is {1, 1, 1, 1, 1, 2, 3}.

   • Case 3: If z = 2 Substituting z=2 into y + 2z = 3 gives y + 2(2) = 3, so y = -1. This is not possible, as the number of digits cannot be negative.

   Any value of z > 1 will also result in a negative y. Therefore, we only have two valid cases.

3. Calculate the number of integers for each case. For each valid set of digits, we need to find the number of distinct permutations. This can be calculated using the multinomial coefficient formula: n! / (n1! * n2! * ... * nk!).

   • For Case 1: The digits are {1, 1, 1, 1, 2, 2, 2}. We have 7 digits in total, with 4 identical 1s and 3 identical 2s. Number of arrangements = $\frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 7 \times 5 = 35$

   • For Case 2: The digits are {1, 1, 1, 1, 1, 2, 3}. We have 7 digits in total, with 5 identical 1s. Number of arrangements = $\frac{7!}{5!1!1!} = \frac{7 \times 6}{1} = 42$

4. Find the total number of integers. The total number of possible seven-digit integers is the sum of the numbers of arrangements from each case, as the cases are mutually exclusive. Total number of integers = (Number from Case 1) + (Number from Case 2) Total number of integers = $35 + 42 = 77$

Thus, there are 77 such seven-digit integers.