Literally too many variables!-2

Given that $a = \dfrac {b+c}{x-2} \implies x = \dfrac {b+c}a + 2 = \dfrac {a+b+c}a + 1 = \alpha + 1$ , where $\alpha =\dfrac {a+b+c}a$ . Similarly, $y = \beta + 1$ and $z = \gamma +1$ . where $\beta =\dfrac {a+b+c}b$ and $\gamma =\dfrac {a+b+c}c$ . Then we have:

$\begin{aligned} x+y+z & = 2010 \\ \alpha + 1 + \beta + 1 + \gamma + 1 & = 2010 \\ \color{#3D99F6} \alpha + \beta + \gamma & \color{#3D99F6} = 2007 & \color{#3D99F6} ...(1) \end{aligned}$

$\begin{aligned} xy+yz+zx & = 67 \\ \sum_{cyc} (\alpha + 1)(\beta + 1) & = 67 \\ \sum_{cyc} (\alpha \beta + \alpha + \beta + 1) & = 67 \\ \alpha \beta + \beta \gamma + \gamma \alpha + 2(\alpha + \beta + \gamma) + 3 & = 67 \\ \alpha \beta + \beta \gamma + \gamma \alpha & = 67 - 2({\color{#3D99F6}\alpha + \beta + \gamma}) - 3 & \small \color{#3D99F6} (1): \quad \alpha + \beta + \gamma = 2007 \\ \color{#D61F06} \alpha \beta + \beta \gamma + \gamma \alpha & \color{#D61F06} = - 3950 & \color{#D61F06} ... (2) \end{aligned}$

Now consider

$\begin{aligned} \color{#D61F06} \alpha \beta + \beta \gamma + \gamma \alpha & = (a+b+c)^2 \left(\frac 1{ab} + \frac 1{bc} + \frac 1{ca} \right) \\ & = (a+b+c)^2 \left(\frac {a+b+c}{abc} \right) \\ & = \frac {(a+b+c)^3}{abc} \\ & = \alpha \beta \gamma \\ \implies \color{#D61F06} \alpha \beta \gamma & \color{#D61F06} = - 3950 & \color{#D61F06} ...(3) \end{aligned}$

At last,

$\begin{aligned} xyz & = (\alpha + 1)(\beta + 1)(\gamma + 1) \\ & = {\color{#D61F06}\alpha \beta \gamma} + {\color{#3D99F6}\alpha + \beta + \gamma} + {\color{#D61F06} \alpha \beta + \beta \gamma + \gamma \alpha} + 1 \\ & = {\color{#D61F06}-3950} + {\color{#3D99F6}2007} + {\color{#D61F06}-3950} + 1 \\ \implies - xyz & = \boxed{5892} \end{aligned}$