13. In $\triangle A B C$, $\angle B=\frac{\pi}{4}, \angle C=\frac{5 \pi}{12}, A C$ $=2 \sqrt{6}, A C$'s midpoint is $D$. If a line segment $P Q$ of length 3 (point $P$ to the left of point $Q$) slides on line $B C$, then the minimum value of $A P+D Q$ is $\qquad$.
13. $\frac{\sqrt{30}+3 \sqrt{10}}{2}$.

Given that $B C=6$.
Draw a line $D E / / B C$ through point $D$, intersecting $A B$ at point $E$.
Then $D E=\frac{1}{2} B C=3$.
Thus, quadrilateral $P Q D E$ is a parallelogram, i.e., $D Q=E P$.

Therefore, the problem is reduced to: finding a point on line $B C$ such that $A P+E P$ is minimized.
The minimum value of $A P+E P$ is calculated to be $\frac{\sqrt{30}+3 \sqrt{10}}{2}$.