Let the seven-digit number contain:

• $x$ digits equal to $1$
• $y$ digits equal to $2$
• $z$ digits equal to $3$

Since the number has $7$ digits, $x+y+z=7$

Since the sum of digits is $10$, $x+2y+3z=10$

Subtract the first equation from the second: $x+2y+3z-(x+y+z)=10-7$ $y+2z=3$

Now solve in non-negative integers: $y+2z=3$

Possible values are:

• If $z=0$, then $y=3$
• If $z=1$, then $y=1$
• $z\ge 2$ is not possible

Case 1: $z=0,\ y=3$

Then $x=7-3-0=4$

So the digits are four $1$'s and three $2$'s. Number of such arrangements: $\frac{7!}{4!3!}=35$

Case 2: $z=1,\ y=1$

Then $x=7-1-1=5$

So the digits are five $1$'s, one $2$, and one $3$. Number of such arrangements: $\frac{7!}{5!1!1!}=42$

Total number of seven-digit integers: $35+42=77$

Therefore, the correct option is $\boxed{77}$ which is option $\boxed{D}$.

Comparison with stored answer:

Stored correct answer is $D$, and our derived answer is also $D$.