14. Let there be a rectangular parallelepiped $A_{1} B_{1} C_{1} D_{1}-A B C D$, with three edges $A A_{1}=a, B_{1} A_{1}=b, A_{1} D_{1}=c, M, N, P, Q$ being the midpoints of $A_{1} B_{1}, A_{1} D_{1}, B C, C D$ respectively. Find the distance between the centroids of the two triangles $A M N$ and $C_{1} P Q$.
14. As shown in the upper right figure, make the section $A A_{1} C_{1} C$. Suppose it intersects $\triangle A M N$ and $\triangle C_{1} P Q$ at $A E_{1}$ and $C_{1} E$ respectively. Let the diagonals of rectangle $A B C D$ intersect at $H$, then $A H=H C=B H=H D$, so $P E=E Q=C E=\frac{1}{4} A C$. Therefore, $C_{1} E$ is the median of $\triangle C_{1} P Q$, and the orthocenter $G$ of $\triangle C_{1} P Q$ must lie on $C_{1} E$. Similarly, $A_{1} E_{1}=\frac{1}{4} A_{1} C_{1}$, and the centroid $G_{1}$ of $\triangle A M N$ must lie on $A E_{1}$.

In the section, establish a coordinate system as shown in the lower right figure. $A_{1}$ is the origin, and the coordinates of $E_{1}$, $C_{1}$, and $E$ are $\left(\frac{1}{4} d, 0\right)$, $(d, 0)$, and $\left(\frac{3}{4} d, a\right)$ respectively. Here, $d=\sqrt{b^{2}+c^{2}}$.

Take $F$ on $C_{1} E$ such that $C_{1} F=\frac{1}{3} C_{1} E$. It is easy to know that $E_{1} F$ is parallel and equal to $G_{1} G$. By the section formula, the coordinates of $F$ are $\left(\frac{11}{12} d, \frac{a}{3}\right)$, so $G_{1} G=E_{1} F=$ $\sqrt{\left(\frac{11}{12} d-\frac{1}{4} d\right)^{2}+\left(\frac{a}{3}\right)^{2}}=\frac{1}{3} \sqrt{a^{2}+4 b^{2}+4 c^{2}}$.