25. Find three prime numbers, the product of which is three times their sum.
25. Let the required numbers be $a, b$ and $c$, then $a b c=3(a+b+c)$, from which $a+b+c=a b c: 3$. The equality is possible under the condition that one of the numbers is equal to 3. Let $a=3$, then $3+b+c=b c$. Of the two prime numbers $b$ and $c$, at least one is odd. Let $b=2 n+1$, then $3+2 n+1+c=2 c n+c$, from which $n+2=c n$ and $n=2:(c-1)$. The last equality is valid for $c=2$, in which case $b=5$. Therefore, the required numbers are $2,3,5$, and then $2 \cdot 3 \cdot 5=3 \cdot(2+3+5)$.