2. Let point $P$ be inside the face $A B C$ of a regular tetrahedron $A B C D$ with edge length 2, and its distances to the faces $D A B$, $D B C$, and $D C A$ form an arithmetic sequence. Then the distance from point $P$ to the face $D B C$ is $\qquad$
2. $\frac{2 \sqrt{6}}{9}$.

From the problem, we know that the required distance is one-third of the sum of the distances from point $P$ to the planes $DAB$, $DBC$, and $DCA$.

The sum of the distances from point $P$ to the planes $DAB$, $DBC$, and $DCA$ is equal to the height of the regular tetrahedron. The height of a regular tetrahedron with edge length 2 is $\frac{2 \sqrt{6}}{3}$, therefore, the required distance is $\frac{2 \sqrt{6}}{9}$.