12. For any positive integer $n$, define the function $\mu(n)$:
$$
\mu(1)=1 \text {, }
$$

and when $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{t}^{\alpha_{4}} \geqslant 2$,
$$
\mu(n)=\left\{\begin{array}{ll}
(-1)^{t}, & \alpha_{1}=\alpha_{2}=\cdots=\alpha_{t}=1 ; \\
0, & \text { otherwise, }
\end{array}\right.
$$

where, $t \geqslant 1, p_{1}, p_{2}, \cdots, p_{t}$ are distinct primes.
If we denote $A=\left\{x_{1}, x_{2}, \cdots, x_{k}\right\}$ as the set of all distinct positive divisors of 12, then $\sum_{i=1}^{k} \mu\left(x_{i}\right)=$ $\qquad$
12.0.

Given $12=2^{2} \times 3$, we know $A=\left\{1,2,3,2^{2}, 2^{2} \times 3\right\}$.
$$
\begin{array}{l} 
\text { Then } \sum_{i=1}^{k} \mu\left(x_{i}\right) \\
= \mu(1)+(\mu(2)+\mu(3))+\mu(2 \times 3)+ \\
\left(\mu\left(2^{2}\right)+\mu\left(2^{2} \times 3\right)\right) \\
= 1+\left[(-1)^{1}+(-1)^{1}\right]+(-1)^{2}+(0+0) \\
= 0 .
\end{array}
$$