7. [5] Let $a, b, c, x, y$, and $z$ be complex numbers such that
$$
a=\frac{b+c}{x-2}, \quad b=\frac{c+a}{y-2}, \quad c=\frac{a+b}{z-2} .
$$

If $x y+y z+z x=67$ and $x+y+z=2010$, find the value of $x y z$.
Answer: -5892 Manipulate the equations to get a common denominator: $a=\frac{b+c}{x-2} \Longrightarrow x-2=$ $\frac{b+c}{a} \Longrightarrow x-1=\frac{a+b+c}{a} \Longrightarrow \frac{1}{x-1}=\frac{a}{a+b+c}$; similarly, $\frac{1}{y-1}=\frac{b}{a+b+c}$ and $\frac{1}{z-1}=\frac{c}{a+b+c}$. Thus
$$
\begin{aligned}
\frac{1}{x-1}+\frac{1}{y-1}+\frac{1}{z-1} & =1 \\
(y-1)(z-1)+(x-1)(z-1)+(x-1)(y-1) & =(x-1)(y-1)(z-1) \\
x y+y z+z x-2(x+y+z)+3 & =x y z-(x y+y z+z x)+(x+y+z)-1 \\
x y z-2(x y+y z+z x)+3(x+y+z)-4 & =0 \\
x y z-2(67)+3(2010)-4 & =0 \\
x y z & =-5892
\end{aligned}
$$