2. Given a rectangular cuboid $A B C D-A_{1} B_{1} C_{1} D_{1}$ with length, width, and height of $1, 2, 3$ respectively, and $P$ is a point within the plane $A_{1} B D$. Then the minimum length of $A P$ is


2. $\frac{6}{7}$.

Notice that, $A P$ is shortest if and only if $A P \perp$ plane $A_{1} B D$. At this time, by the Pythagorean theorem, we get $A_{1} D=\sqrt{5}, A_{1} B=\sqrt{10}, B D=\sqrt{13}$. Then $\cos \angle B A_{1} D=\frac{\sqrt{2}}{10} \Rightarrow \sin \angle B A_{1} D=\frac{7 \sqrt{2}}{10}$. Thus, $S_{\triangle A, B D}=\frac{7}{2}$. Therefore, $A P_{\min }=\frac{6}{7}$.