12. (5 points) As shown in the figure, the front of each of the three cards has a number, and their backs are written with prime numbers $m, n, p$. If the sum of the numbers on both sides of each card is equal, then the minimum value of $m+n+p$ is $\qquad$ .
【Solution】Solution: Let the sum of the two numbers be $k$, then according to the problem, we have,
$$
47+m=53+n=71+p=k,
$$

Thus, $m=k-47, n=k-53, p=k-71$,
Therefore, $m+n+p=3k-(47+53+71)$,
Since $m, n, p$ are all prime numbers, and cannot be equal to 2, $m, n, p$ must all be odd primes. Assuming $p=3$, then $n=71+3-53=21, 21$ is a composite number, which does not meet the problem's requirements. Assuming $p=5$, then $n=71+5-53=23, 23$ is a prime number, which meets the problem's requirements, and $m=71+5-47=29, 29$ is a prime number, which meets the problem's requirements,
Therefore, the minimum value of $m+n+p$ is: $29+23+5=57$. Answer: The minimum value of $m+n+p$ is 57.
Hence, the answer is: 57.