Let non-zero real numbers $a, b, c$ satisfy $a+b+c\ne 0$. If the following relations hold $$\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}$$ Find the value of $$\frac{(a+b)(b+c)(c+a)}{abc}$$