jmc

We have that $$x-2=\frac{b+c}{a},y-2=\frac{a+c}{b},z-2=\frac{a+b}{c},$$so $$x-1=\frac{a+b+c}{a},y-1=\frac{a+b+c}{b},z-1=\frac{a+b+c}{c}.$$Then $$\frac{1}{x-1}=\frac{a}{a+b+c},\frac{1}{y-1}=\frac{b}{a+b+c},\frac{1}{z-1}=\frac{c}{a+b+c},$$so $$\frac{1}{x-1}+\frac{1}{y-1}+\frac{1}{z-1}=\frac{a+b+c}{a+b+c}=1.$$Multiplying both sides by $(x-1)(y-1)(z-1),$ we get $$(y-1)(z-1)+(x-1)(z-1)+(x-1)(y-1)=(x-1)(y-1)(z-1).$$Expanding, we get $$xy+xz+yz-2(x+y+z)+3=xyz-(xy+xz+yz)+(x+y+z)-1,$$so $$xyz=2(xy+xz+yz)-3(x+y+z)+4=2\cdot 5-3\cdot 3+4=\boxed{5}.$$