II. (50 points)
Solve the system of equations: $\left\{\begin{array}{l}a b+a+2 b=78 \\ b c+3 b+c=101 \\ c d+5 c+3 d=232 \\ d e+4 d+5 e=360 \\ e a+2 e+4 a=192\end{array}\right.$
The original system of equations can be transformed into $\left\{\begin{array}{l}(a+2)(b+1)=80 \\ (b+1)(c+3)=104 \\ (c+3)(d+5)=247 \\ (d+5)(e+4)=380 \\ (e+4)(a+2)=200\end{array}\right.$
(1) $\times$ (3) $\times$ (5) $\div$ (2) $\div$ (4) gives $(a+2)^{2}=100$ so $a+2= \pm 10$.

Substituting sequentially, we get $b+1= \pm 8, c+3= \pm 13, d+5= \pm 19, e+4= \pm 20$.
Therefore, $(a, b, c, d, e)=(8,7,10,14,16)$ or $(-12,-9,-16,-24,-24)$.