7. Given that $P$ is a point outside the plane $\alpha$ where $\triangle A B C$ lies, $P A \perp$ plane $\alpha, P B=P C=\sqrt{26}, \tan \angle P B C=\frac{3}{2}$. Then the maximum value of the distance from point $A$ to plane $P B C$ is $\qquad$
7. $\frac{3 \sqrt{2}}{2}$.

As shown in Figure 2, draw $P D \perp B C$ at point $D$, connect $A D$, and draw $A F \perp P D$ at point $F$.
Since $P A \perp$ plane $A B C, B C \perp P D$, we have
$B C \perp A D \Rightarrow B C \perp$ plane $P A D$
$\Rightarrow$ plane $P A D \perp$ plane $P B C$.
From $A F \perp P D$, we get $A F \perp$ plane $P B C$.
Thus, $A F$ is the distance from point $A$ to plane $P B C$.
Since $\tan \angle P B C=\frac{3}{2}$, we have
$\sin \angle P B D=\frac{3}{\sqrt{13}}$.
Also, $P D=P B \sin \angle P B D=\sqrt{26} \times \frac{3}{\sqrt{13}}=3 \sqrt{2}$.
In the right triangle $\triangle P A D$,
$A F \leqslant \frac{1}{2} P D=\frac{3 \sqrt{2}}{2}$.
Therefore, the maximum value sought is $\frac{3 \sqrt{2}}{2}$.