9. In a regular tetrahedron $P-ABC$, $M$ is a moving point within $\triangle ABC$ (including the boundary), and the distances from point $M$ to the three lateral faces $PAB$, $PBC$, and $PCA$ form an arithmetic sequence. Then the trajectory of point $M$ is ( ).
(A) a broken line segment
(B) a line segment
(C) a segment of a circle
(D) a segment of a parabola
9 As shown in the figure, since the three side areas of the regular tetrahedron $P-ABC$ are equal, the distances from point $M$ to the three side faces $PAB$, $PBC$, and $PCA$ form an arithmetic sequence, which is equivalent to the volumes of the three tetrahedrons $M-PAB$, $M-PBC$, and $M-PCA$ forming an arithmetic sequence, i.e.,
$$
2 V_{M-PBC} = V_{M-PAB} + V_{M-PCA}.
$$

Therefore, $3 V_{M-PBC} = V_{P-ABC}$, which means $V_{P-MBC} = \frac{1}{3} V_{P-ABC}$, thus $S_{\triangle MBC} = \frac{1}{3} S_{\triangle ABC}$. Hence, the locus of point $M$ is a segment parallel to $BC$ passing through the centroid of $\triangle ABC$. Therefore, the answer is $\mathrm{B}$.