** Given in the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, the edge lengths are $A B=B C=3$, $B B_{1}=4$. Connect $B_{1} C$, and draw a perpendicular from point $B$ to $B_{1} C$ intersecting $C C_{1}$ at point $E$ and $B_{1} C$ at point $F$.
(1) Prove: $A_{1} C \perp$ plane $E B D$;
(2) Let $A_{1} C \cap$ plane $E B D=K$, find the length of segment $A_{1} K$.
(1) Taking $D$ as the origin, the lines $D A$, $D C$, and $D D_{1}$ as the $x$-axis, $y$-axis, and $z$-axis respectively, we establish a spatial rectangular coordinate system. Then, $D(0,0,0)$, $A(3,0,0)$, $C(0,3,0)$, $B(3,3,0)$, $A_{1}(3,0,4)$, $D_{1}(0,0,4)$, $C_{1}(0,3,4)$, and $B_{1}(3,3,4)$. Let $E(0,3, z)$, then $\overrightarrow{B E}=(-3,0, z)$, $\overrightarrow{B_{1} C}=(-3,0,-4)$. From $\overrightarrow{B E} \cdot \overrightarrow{B_{1} C}=0$, we have $\overrightarrow{B E} \cdot \overrightarrow{B_{1} C}=(-3,0, z) \cdot(-3,0,-4)$ $=9-4 z=0$. Thus, $z=\frac{9}{4}$. Therefore, $E\left(0,3, \frac{9}{4}\right)$. Hence, $\overrightarrow{A_{1} C} \cdot \overrightarrow{D B}=-3 \times 3+3 \times 3=$ $0, \overrightarrow{A_{1} C} \cdot \overrightarrow{D E}=3 \times 3-4 \times \frac{4}{a}=0$. That is, $\overrightarrow{A_{1} C} \perp \overrightarrow{D B}$, $\overrightarrow{A_{1} C} \perp \overrightarrow{D E}$. Therefore, $A_{1} C \perp D B$, $A_{1} C \perp D E$. Consequently, $A_{1} C \perp$ plane $E B D$.
(2) From $\overrightarrow{D K}=m \overrightarrow{D B}+n \overrightarrow{D E}=m(3,3,0)+$ $n\left(0,3, \frac{9}{4}\right)=\left(3 m, 3 m+3 n, \frac{9}{4} n\right)$, we know $K(3 m, 3 m+$ $\left.3 n, \frac{9}{4} n\right)$. Therefore, $\overrightarrow{A_{1} K}=\left(3 m-3,3 m+3 n, \frac{9}{4} n-4\right)$. Since $\overrightarrow{A_{1} K} \perp \overrightarrow{D B}$, we have $\overrightarrow{A_{1} K} \cdot \overrightarrow{D B}=\left(3 m-3,3 m+3 n, \frac{9}{4} n-4\right) \cdot$ $(3,3,0)=0$, which gives $2 m+n-1=0$. (1) Similarly, from $\overrightarrow{A_{1} K} \perp$ $\overrightarrow{D E}$, we have $16 m+25 n-16=0$. (2) Solving equations (1) and (2) yields $m=\frac{9}{34}, n=\frac{8}{17}$. Therefore, $\overrightarrow{A_{1} K}=$ $\left(-\frac{75}{34}, \frac{75}{34},-\frac{50}{17}\right)$. Hence, $\left|\overrightarrow{A_{1} K}\right|=\frac{25 \sqrt{34}}{34}$ is the length of $A_{1} K$.