6. Given that $P$ is a point on a sphere $O$ with radius $r$, and three mutually perpendicular moving chords $P A$, $P B$, and $P C$ are drawn through $P$. If the maximum distance from point $P$ to the plane $A B C$ is 1, then $r=$ $\qquad$
6. $\frac{3}{2}$.

Let $P A=x, P B=y, P C=z$. Then
$$
\begin{array}{l}
A B^{2}=x^{2}+y^{2}, B C^{2}=y^{2}+z^{2}, C A^{2}=z^{2}+x^{2}, \\
S_{\triangle A B C}=\frac{1}{2} A B \cdot B C \sin \angle A B C \\
=\frac{1}{2} A B \cdot B C \sqrt{1-\cos ^{2} \angle A B C} \\
=\frac{1}{2} \sqrt{x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}, \\
V_{\text {tetrahedron } P-A B C}=\frac{1}{6} x y z=\frac{1}{3} h S_{\triangle A B C} .
\end{array}
$$

Therefore, $h=\frac{x y z}{\sqrt{x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}}=\frac{1}{\sqrt{\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}}}$
$$
\leqslant \sqrt{\frac{x^{2}+y^{2}+z^{2}}{9}}=\sqrt{\frac{4 r^{2}}{9}} .
$$

Thus, $r=\frac{3}{2}$.