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 $\sum_{k=1}^{n}\left(x_{k}+y_{k}\right)=$
$\qquad$ .
$$
x y-6(x+y)+36=36 \Rightarrow(x-6)(y-6)=1 \cdot 36=2 \cdot 18=3 \cdot 12=4 \cdot 9=6 \cdot 6 \text {, }
$$

Then $(x, y)=(7,42),(8,24),(9,18),(10,15),(12,12),(15,10),(18,9),(24,8),(42,7)$. So $\sum_{k=1}^{n}\left(x_{k}+y_{k}\right)=2(49+32+27+25)+24=290$.