If $\left|\begin{array}{ccc}a+b& b+c& c+a\\ b+c& c+a& a+b\\ c+a& a+b& b+c\end{array}\right|=K\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$, then $K=$

Let ${\Delta }_1=\left|\begin{array}{ccc}a+b& b+c& c+a\\ b+c& c+a& a+b\\ c+a& a+b& b+c\end{array}\right|$.Using property of determinants:
$${\Delta }_1=\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|+\left|\begin{array}{ccc}b& c& a\\ c& a& b\\ a& b& c\end{array}\right|$$
By interchanging columns (each interchange multiplies the determinant by $-1$):In the second determinant, swap $C_1$ and $C_2$, then swap $C_2$ and $C_3$. Two swaps result in $(-1{)}^2=1$.
$$\left|\begin{array}{ccc}b& c& a\\ c& a& b\\ a& b& c\end{array}\right|=\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$$
Thus, ${\Delta }_1=2{\Delta }_2$.Therefore, $K=2$.