12. Given real numbers $a, b, c$ satisfy
$$
\frac{a(b-c)}{b(c-a)}=\frac{b(c-a)}{c(b-a)}=k>0,
$$

where $k$ is some constant. Then the greatest integer not greater than $k$ is . $\qquad$
12. 0 .

Let the substitution be $x=a b, y=b c, z=c a$. Then
$$
x-z=k(y-x), y-x=k(y-z) \text {. }
$$

Thus, $x-z=k^{2}(y-z)$. Therefore,
$$
\begin{array}{l}
(y-x)+(x-z)+(z-y)=0 \\
\Leftrightarrow(y-z)\left(k^{2}+k-1\right)=0(x \neq y \neq z) \\
\Rightarrow k^{2}+k-1=0 .
\end{array}
$$

Also, $k>0 \Rightarrow k=\frac{-1+\sqrt{5}}{2} \Rightarrow[k]=0$.