3. (3 points) There are 4 numbers of different sizes. If any 3 numbers are taken to add up, the sums obtained are $180, 197, 208$, and 222. Then, the sum that definitely does not include the second smallest number is ( )
A. 180
B. 197
C. 208
D. 222
【Analysis】Let the four different numbers be $a, b, c, d$. According to the problem, $(a+b+c)+(a+c+d)+(b+c+d)$ $+(a+b+d)=3(a+b+c+d)=180+197+208+222=807$, then $a+b+c+d=269$. From this, we can determine which sum does not include the second smallest number.

【Solution】Let the four different numbers be $a, b, c, d$. Then
$$
\begin{aligned}
& (a+b+c)+(a+c+d)+(b+c+d)+(a+b+d) \\
= & 3(a+b+c+d) \\
= & 180+197+208+222 \\
= & 807
\end{aligned}
$$

Therefore, $a+b+c+d=807 \div 3=269$.
The smallest number should be: $269-222=47$,
The second smallest number is: $269-208=61$.
Thus, the sum that does not include the second smallest number must be 208.
Hence, the answer is: $C$.