6. In $\triangle A B C$ with side lengths $8, 10, 12$, there is a point $P$. Perpendiculars from $P$ to $A B, B C, C A$ meet these sides at $D, E, F$ respectively. When $A D^{2}+B E^{2}+C F^{2}$ reaches its minimum value of 77, $P$ is the ( ) of $\triangle A B C$.
(A) Incenter
(B) Circumcenter
(C) Centroid
(D) Orthocenter
6. B.
$$
\text { Let } B C=12, A C=10, A B=8, B E=x, C F
$$
$=y, A D=z$. As shown in Figure 4.
By the Pythagorean theorem, we have
$$
x^{2}+y^{2}+z^{2}=(12-x)^{2}+(10-y)^{2}+(8-z)^{2} \text {. }
$$

Let $x \leqslant 6$, i.e., $x=6-x_{1}\left(x_{1} \geqslant 0\right)$. Then
$$
\begin{array}{l}
x^{2}+(12-x)^{2}=\left(6-x_{1}\right)^{2}+\left(6+x_{1}\right)^{2} \\
=72+2 x_{1}^{2} .
\end{array}
$$

Similarly, $y^{2}+(10-y)^{2}=50+2 y_{1}^{2}$,
$$
z^{2}+(8-z)^{2}=32+2 z_{1}^{2} \text {. }
$$

Thus, $x^{2}+y^{2}+z^{2}$
$$
\begin{aligned}
= & \frac{1}{2}\left[x^{2}+(12-x)^{2}+y^{2}+(10-y)^{2}+\right. \\
& \left.z^{2}+(8-z)^{2}\right] \\
= & 77+x_{1}^{2}+y_{1}^{2}+z_{1}^{2}=77 \\
\Rightarrow & x_{1}=y_{1}=z_{1}=0
\end{aligned}
$$
$\Rightarrow P$ is the circumcenter of $\triangle A B C$.