12. In a regular tetrahedron $ABCD$, $MN$ is the line segment connecting the midpoint of $AD$ and the center of $\triangle BCD$, and $PQ$ is the line segment connecting the midpoint of $CD$ and the center of $\triangle ABC$. Find the angle formed by the lines $MN$ and $PQ$.
Three, 12. As shown in the figure, let the edge length of the regular tetrahedron $ABCD$ be $a$, and $M, P$ be the midpoints of $AD$ and $DC$ respectively. $N, Q$ are the centers of $\triangle BCD$ and $\triangle ABC$ respectively. Take the midpoint $F$ of $BC$, and connect $AF, DF$, knowing that $AF=DF=\frac{\sqrt{3}}{2} a$. $Q, H$ are on $AF, DF$ respectively, and $QE \parallel NM$ intersects $AD$ at $E$, $\angle PQE$ is the angle formed by $MN$ and $PQ$ (or $\pi-\angle PQE$). Also, $FQ=FN=\frac{1}{3} DF=\frac{\sqrt{3}}{b} a$. Connect $PE, NQ$, since $NQ \parallel AD$, therefore $MNQE$ is a parallelogram, thus $NQ=\frac{1}{3} a, DE=\frac{5}{6} a$. Connect $AN$, then $AN \perp$ plane $BCD$, thus $AN \perp DF$, thus $MN=\frac{1}{2} a$. Similarly, $PQ=\frac{1}{2} a$. In $\triangle DPE$, $PE^{2}=DE^{2}+DP^{2}-2 DE \cdot DP \cdot \cos \angle EDP=\left(\frac{5}{6} a\right)^{2}+\left(\frac{1}{2} a\right)^{2}-2 \cdot \frac{5}{6} a \cdot \frac{1}{2} a \cdot \frac{1}{2}=\frac{19}{36} a^{2}$. In $\triangle PQE$, $\cos \angle PQE=\frac{EQ^{2}+PQ^{2}-EP^{2}}{2 EQ \cdot PQ}=-\frac{1}{18}$, therefore the angle formed by $MN, PQ$ is $\arccos \frac{1}{18}$.