12. Let $a=\lg z+\lg \left[x(y z)^{-1}+1\right], b=\lg x^{-1}+\lg (x y z+1), c=\lg y+\lg \left[(x y z)^{-1}+1\right]$, and let the maximum of $a, b, c$ be $M$. Then the minimum value of $M$ is $\qquad$.
$$
a=\lg \left(\frac{x}{y}+z\right), b=\lg \left(y z+\frac{1}{x}\right), c=\lg \left(\frac{1}{x z}+y\right) .
$$

Let $M_{1}=\max \left\{\frac{x}{y}+z, y z+\frac{1}{x}, \frac{1}{x z}+y\right\}$, then $2 M_{1} \geqslant \frac{x}{y}+z+\frac{1}{x z}+y \geqslant 4 \sqrt[4]{\frac{x}{y} \cdot z \cdot \frac{1}{x z} \cdot y}$ $=4 \Rightarrow M_{1} \geqslant 2$, equality holds when $\frac{x}{y}=z=\frac{1}{x z}=y=1 \Rightarrow x=y=z=1$. At this point, $y z+\frac{1}{x}$ also equals 2, so the minimum value of $M$ is $\lg 2$.