14. As shown in the figure, $P-A B C$ is a regular triangular pyramid with a base edge length of 1. $D, E, F$ are points on the edges $P A, P B, P C$ respectively, and the section $D E F \parallel$ the base $A B C$. The sum of the edge lengths of the frustum $D E F-A B C$ is equal to the sum of the edge lengths of the pyramid $P-A B C$. The sum of the edge lengths refers to the total length of all edges in the polyhedron.
(1) Prove that $P-A B C$ is a regular tetrahedron;
(2) If $P D=\frac{1}{2} P A$, find the size of the dihedral angle $D-B C-A$:
(3) Let the volume of the frustum $D E F-A B C$ be $V$. Does there exist a right parallelepiped with volume $V$ and all edges of equal length such that it has the same sum of edge lengths as the frustum $D E F-A B C$? If it exists, construct such a right parallelepiped and provide a proof; if not, explain the reason.
14. (1) Since the sum of the edge lengths of the frustum $D E F-A B C$ is equal to that of the pyramid $P-A B C$,

Therefore, $D E+E F+F D=P D+P E+P F$.
Also, because the section $D E F \parallel$ base $A B C$,
So $D E=E F=F D=P D=P E=P F, \angle D P E=\angle E P F=$ $\angle F P D=60^{\circ}$,
Therefore, $P-A B C$ is a regular tetrahedron.
(2) Take the midpoint $M$ of $B C$, and connect $P M, D M, A M$.
Since $B C \perp P M, B C \perp A M$, it follows that $B C \perp$ plane $P A M$, thus $B C \perp D M$, so $\angle D M A$ is the plane angle of the dihedral angle $D-B C-A$.
From (1), we know that the edge lengths of $P-A B C$ are all 1, so $P M=A M=\frac{\sqrt{3}}{2}$,
By the given condition, $D$ is the midpoint of $P A$, in the right triangle $\triangle A D M$, we have $\sin \angle D M A=\frac{A D}{A M}=\frac{\sqrt{3}}{3}$,
Therefore, $\angle D M A=\arcsin \frac{\sqrt{3}}{3}$ is the size of the required dihedral angle.
(3) There exists a right parallelepiped that meets the conditions.
The sum of the edge lengths of the frustum $D E F-A B C$ is a constant 6, and the volume is $V$.
Assume the edge lengths of the right parallelepiped are all $\frac{1}{2}$, and the angle between adjacent sides of the base is $\alpha$, then the sum of the edge lengths of this parallelepiped is 6, and the volume is $V=\frac{1}{8} \sin \alpha$.
Since the volume of the regular tetrahedron $P-A B C$ is $\frac{1}{3} \times \frac{\sqrt{3}}{4} \times 1^{2} \times \frac{\sqrt{6}}{3}=\frac{\sqrt{2}}{12}$,
Therefore, $0<V<\frac{\sqrt{2}}{12}$, hence $0<8 V<1$. It follows that $\alpha=\operatorname{arcsin}(8 V)$, so constructing a right parallelepiped with edge lengths all $\frac{1}{2}$ and the angle between adjacent sides of the base being $\arcsin (8 V)$ meets the requirements.