6. Let $A B C$ be a triangle with $A B=5, B C=6, C A=7$. Let $D$ be a point on ray $A B$ beyond $B$ such that $B D=7, E$ be a point on ray $B C$ beyond $C$ such that $C E=5$, and $F$ be a point on ray $C A$ beyond $A$ such that $A F=6$. Compute the area of the circumcircle of $D E F$.
Answer:
$\frac{251}{3} \pi$
Solution 1: Let $I$ be the incenter of $A B C$. We claim that $I$ is the circumcenter of $D E F$.
To prove this, let the incircle touch $A B, B C$, and $A C$ at $X, Y$, and $Z$, respectively. Noting that $X B=B Y=2, Y C=C Z=4$, and $Z A=A X=3$, we see that $X D=Y E=Z F=9$. Thus, since $I X=I Y=I Z=r$ (where $r$ is the inradius) and $\angle I X D=\angle I Y E=\angle I Z F=90^{\circ}$, we have three congruent right triangles, and so $I D=I E=I F$, as desired.
Let $s=\frac{5+6+7}{2}=9$ be the semiperimeter. By Heron's formula, $[A B C]=\sqrt{9(9-5)(9-6)(9-7)}=$ $6 \sqrt{6}$, so $r=\frac{[A B C]}{s}=\frac{2 \sqrt{6}}{3}$. Then the area of the circumcircle of $D E F$ is
$$
I D^{2} \pi=\left(I X^{2}+X D^{2}\right) \pi=\left(r^{2}+s^{2}\right) \pi=\frac{251}{3} \pi .
$$

Solution 2: Let $D^{\prime}$ be a point on ray $C B$ beyond $B$ such that $B D^{\prime}=7$, and similarly define $E^{\prime}, F^{\prime}$. Noting that $D A=E^{\prime} A$ and $A F=A F^{\prime}$, we see that $D E^{\prime} F^{\prime} F$ is cyclic by power of a point. Similarly, $E F^{\prime} D^{\prime} D$ and $F D^{\prime} E^{\prime} E$ are cyclic. Now, note that the radical axes for the three circles circumscribing these quadrilaterals are the sides of $A B C$, which are not concurrent. Therefore, $D D^{\prime} F F^{\prime} E E^{\prime}$ is cyclic. We can deduce that the circumcenter of this circle is $I$ in two ways: either by calculating that the midpoint of $D^{\prime} E$ coincides with the foot from $I$ to $B C$, or by noticing that the perpendicular bisector of $F F^{\prime}$ is $A I$. The area can then be calculated the same way as the previous solution.