$174 S-A B C D-S^{\prime}$ is a regular octahedron with edge length 1, and sphere $O$ is outside of it, tangent to face $S A B$, and also tangent to the extended planes of three faces $S B C, S A D, S^{\prime} A B$, then the radius $r=$ $\qquad$
$174 \frac{1}{4} \sqrt{\frac{2}{3}}$. Draw a line $S P$ through $S$ parallel to $A D$, intersecting the plane $S^{\prime} A B$ (extended) at $P$.
Since $A D / / B C$, the intersection line of planes $S A D$ and $S B C$ is parallel to $A D$, and thus is $S P$. $P A$ and $P B$ are the intersection lines of plane $S^{\prime} A B$ with planes $S A D$ and $S B C$, respectively.
$S S^{\prime}$ passes through the center $O$ of the square $A B C D$. Let $P S^{\prime}$ intersect $A B$ at $E$, then because $S P / / A B$, we have $O E / / S P / / A D \cdot S S^{\prime}=2 O S^{\prime}$, so $S P=2 O E=A B=1$. Also, $\angle P S B=60^{\circ}$, so $\triangle P S B$ is an equilateral triangle. Similarly, $\triangle P S A$ is an equilateral triangle. Therefore, $\triangle P A B$ is an equilateral triangle. The tetrahedron $P A B S$ is a regular tetrahedron. Its inscribed sphere radius $r=\frac{1}{4} h=\frac{1}{4} \sqrt{\frac{2}{3}}$.