12. (6 points) In a residential building, each household subscribes to 2 different newspapers. If the households in the building only subscribe to newspapers A, B, and C, with 30 copies of newspaper A, 34 copies of newspaper B, and 40 copies of newspaper C. Then the number of households that subscribe to both newspaper B and newspaper C is $\qquad$.
【Analysis】According to the problem, a total of $30+34+40=104$ copies of newspapers A, B, and C were subscribed. It is known that on average, each household subscribed to 2 different newspapers, so the total number of households in the building is $104 \div 2=52$. The households that subscribed to both newspaper B and C are those that did not subscribe to newspaper A. Given that 30 copies of newspaper A were subscribed, the number of households that did not subscribe to newspaper A is $52-30=22$.

【Solution】According to the analysis of the problem, we get:
$$
\begin{aligned}
& (30+34+40) \div 2-30, \\
= & 104 \div 2-30, \\
= & 52-30, \\
= & 22 \text { (households); }
\end{aligned}
$$

Answer: 22 households subscribed to both newspaper B and C.
Therefore, the answer is: 22.