Three. (50 points) There is an $m \times n \times p$ rectangular box, and another $(m+2) \times(n+2) \times(p+2)$ rectangular box, where $m, n, p$ are positive integers ($m \leqslant n \leqslant p$), and the volume of the former is half of the latter. Find the maximum value of $p$.

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Three. (50 points) There is an $m \times n \times p$ rectangular box, and another $(m+2) \times(n+2) \times(p+2)$ rectangular box, where $m, n, p$ are positive integers ($m \leqslant n \leqslant p$), and the volume of the former is half of the latter. Find the maximum value of $p$.
Three, Solution: From the problem, $2 m n p=(m+2)(n+2)(p+2)$, we get $\left(1+\frac{2}{m}\right)\left(1+\frac{2}{n}\right)\left(1+\frac{2}{p}\right)=2$.
(1) When $m \geqslant 8$, by $m \leqslant n \leqslant p$, then $\left(1+\frac{2}{m}\right)\left(1+\frac{2}{n}\right)\left(1+\frac{2}{p}\right) \leqslant\left(1+\frac{2}{8}\right)^{3}2$, contradiction!
(3) When $m=3$, then $6 n p=5(n+2)(p+2)$, i.e., $(n-10)(p-10)=120$.
So the maximum value of $p$ is 130;
(4) When $m=4$, then $4 n p=3(n+2)(p+2)$, i.e., $(n-6)(p-6)=48$.
So the maximum value of $p$ is 54;
(5) When $m \geqslant 5$, $\left(1+\frac{2}{p}\right)=\frac{2}{\left(1+\frac{2}{m}\right)\left(1+\frac{2}{n}\right)}>\frac{2}{\left(1+\frac{2}{5}\right)\left(1+\frac{2}{5}\right)}$, we get $p<98$.
In summary, the maximum value of $p$ is 130.