1. **State the problem:** Given the universal set $U = \{6, 7, 8, 9, 10, 11, 12\}$ and subset $A = \{7, 8, 11, 12\}$, find the complement $A'$. 2. **Recall the definition:** The complement $A'$ consists of all elements in $U$ that are *not* in $A$. 3. **Apply the definition:** Identify elements in $U$ that are not in $A$: $$A' = U \setminus A = \{6, 7, 8, 9, 10, 11, 12\} \setminus \{7, 8, 11, 12\}$$ 4. **Remove elements of $A$ from $U$:** $$A' = \{6, \cancel{7}, \cancel{8}, 9, 10, \cancel{11}, \cancel{12}\} = \{6, 9, 10\}$$ 5. **Final answer:** $$\boxed{A' = \{6, 9, 10\}}$$ This means the complement of $A$ in $U$ is the set containing 6, 9, and 10.