In a rectangular paper piece $ABCD$, $AB=13\sqrt{3}, BC=12\sqrt{3}, AC, BD$ intersect at point $P$. If $\triangle PBC$ is cut out, and $\triangle ABP, \triangle CDP$ are folded up so that $BP, CP$ coincide, forming a triangular pyramid, then the volume of this triangular pyramid is . $\qquad$
50 594. $V=\frac{1}{3} S h, S=\frac{1}{4} a b, h=\left|B^{\prime} P\right| \cdot \sin \alpha\left(\right.$ After folding, $B, C$ coincide at point $B^{\prime}$, $B^{\prime} P$ forms an angle $\alpha$ with plane $A D P$, it is easy to find $\cos \alpha=\frac{3 a^{2}+b^{2}}{a \sqrt{a^{2}+b^{2}}}$, hence $V=594$.