12. Let the equation $x y=6(x+y)$ have all positive integer solutions $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \cdots,\left(x_{n}, y_{n}\right)$. Then
$$
\sum_{k=1}^{n}\left(x_{k}+y_{k}\right)=
$$
$\qquad$
12. 290 .

The original equation can be transformed into
$$
(x-6)(y-6)=6^{2}=2^{2} \times 3^{2} \text {. }
$$

Note that, the number of positive divisors of $2^{2} \times 3^{2}$ is
$$
(1+2)(1+2)=9
$$

and
$$
\begin{array}{l}
\quad \sum_{k=1}^{9}\left(\left(x_{k}-6\right)+\left(y_{k}-6\right)\right) \\
\quad=2 \sum_{k=1}^{9}\left(x_{k}-6\right)=2 \times \frac{2^{3}-1}{2-1} \times \frac{3^{3}-1}{3-1} \\
=182 . \\
\text { Therefore, } \sum_{k=1}^{9}\left(x_{k}+y_{k}\right)=182+2 \times 9 \times 6=290 .
\end{array}
$$