$22 \cdot 6$ Given three planes $\alpha, \beta, \gamma$, the angle between each pair of planes is $\theta$, and $\alpha \cap \beta=a, \beta \cap \gamma=b, \gamma \cap \alpha=c$. If
Proposition A: $\theta>\frac{\pi}{3}$.
Proposition B: $a, b, c$ intersect at one point.

Then the relationship between these two propositions is
(A) A is a sufficient condition for B but not necessary.
(B) A is a necessary condition for B but not sufficient.
(C) A is a necessary and sufficient condition for B.
(D) None of (A), (B), (C) is correct.
(China High School Mathematics League, 1988)
[Solution] For three planes $\alpha, \beta, \gamma$, there are two possibilities for their pairwise intersection lines $a, b, c$: either $a, b, c$ are pairwise parallel, or $a, b, c$ intersect at a single point.

If proposition A holds, then $a, b, c$ cannot be pairwise parallel. In fact, if $a / / b, \quad b / / c$, then a right prism can be constructed such that its lateral faces lie on the planes $\alpha, \beta, \gamma$. The base triangle of this prism has each interior angle as the dihedral angle between two of the planes $\alpha, \beta, \gamma$, meaning each interior angle of the triangle is $\theta$, and $\theta > \frac{1}{3} \pi$, which is impossible.

This proves that if proposition A holds, then $a, b, c$ intersect at a single point, i.e., A is a sufficient condition for B.

If proposition B holds, as shown in the figure, let $a, b, c$ intersect at $O$. Since the dihedral angles between the planes $\alpha, \beta, \gamma$ are all $\theta$, a tetrahedron $O-ABC$ can be constructed such that the base $\triangle ABC$ is an equilateral triangle, and the lateral faces of the tetrahedron lie on the planes $\alpha, \beta, \gamma$. A plane $ABD \perp OC$ (i.e., the line $c$) is drawn through $AB$, intersecting $OC$ at $D$.
Since $BC$ is the hypotenuse of the right triangle $BCD$, $BC > BD$.
Similarly, $AC > AD$, so in $\triangle ABD$, $AB > AD, AB > BD$,
thus $\angle ADB > \frac{1}{3} \pi$. By the construction, $\angle ADB = \theta$, i.e., $\theta > \frac{1}{3} \pi$.
Therefore, proposition A holds. That is, A is a necessary condition for B.
Hence, the answer is (C).
Note: This solution is based on the point that "two intersecting planes form only one dihedral angle."
However, if it is believed that "two intersecting planes form four dihedral angles, each less than $180^{\circ}$," then the answer to this problem should be (A): A is a sufficient condition for B but not necessary. The proof is as follows:

For three planes $\alpha, \beta, \gamma$ intersecting pairwise, the intersection lines are $a, b, c$, then there are and only two possibilities: $a / / b / / c$, or $a, b, c$ intersect at a single point.
Sufficiency: Each pair of planes intersect, forming equal dihedral angles, which are "complementary" pairs $\theta, \theta'$. When $\theta > \frac{\pi}{3}$ and $\theta' > \frac{\pi}{3}$, i.e., $\theta, \theta'$ are both between $\frac{\pi}{3}$ and $\frac{2\pi}{3}$, $a, b, c$ intersect at a single point.
If $a, b, c$ do not intersect at a single point, then $a / / b / / c$, as shown in the figure. $\alpha, \beta, \gamma$ form a triangular prism, and a cross-section $ABC$ is made intersecting the three lines at points $A$, $B$, $C$. Then the three interior angles and exterior angles of $\triangle ABC$ are the plane angles of the dihedral angles formed by the pairwise intersections of $\alpha, \beta, \gamma$. By the problem statement, the angles are all $\theta$.
So, $\angle ABC = \angle BCA = \angle CAB = \theta = \frac{\pi}{3}$.
The other set (exterior angles) $\theta' = \frac{2\pi}{3}$, which contradicts $\theta, \theta' > \frac{\pi}{3}$.
Therefore, the three lines must intersect at a single point.

Conversely, the necessity is not necessarily true, i.e., when $a, b, c$ intersect at a single point, it is not necessarily true that $\theta, \theta'$ are both greater than $\frac{\pi}{3}$.
As shown in the figure, the three planes $\alpha, \beta, \gamma$ intersect pairwise, with intersection lines $a, b, c$ intersecting at a single point $O$, forming a trihedral angle $O-ABC$. When each face angle $\angle AOB = \angle BOC = \angle COA = \varphi = \arccos \left(-\frac{1}{3}\right)$, it is easy to prove:

The dihedral angles $B-OA-C = B-OC-A = A-OB-C$.
The corresponding other set of dihedral angles must also be equal.
Cut $OA = OB = OC = 1$, forming the tetrahedron $O-ABC$.
The midpoints of $AB, AC$ are $D, E$, and $DF \perp OA$ intersects at $F$, connecting $EF, DE$.
By $\triangle AOB \cong \triangle AOC$, it is easy to prove $EF \perp OA$.
$\angle DFE = \theta$ is the plane angle of the dihedral angle $B-OA-C$.
Thus,
$$
AB = AC = 2 \sin \frac{\varphi}{2},
$$
$$
\begin{array}{l}
AD = AE = DE = \sin \frac{\varphi}{2}, \\
DF = EF = \cos \frac{\varphi}{2} \cdot \sin \frac{\varphi}{2} = \frac{1}{2} \sin \varphi .
\end{array}
$$

By the cosine rule,
$$
\begin{aligned}
\cos \theta & = \frac{DF^2 + EF^2 - DE^2}{2 DF \cdot EF} = \frac{2 DF^2 - DE^2}{2 DF^2} \\
& = 1 - \frac{\sin^2 \frac{\varphi}{2}}{\frac{1}{2} \sin^2 \varphi} = -\frac{1}{2}, \\
\therefore \theta & = \frac{2}{3} \pi, \text{ and the corresponding } \theta' = \frac{\pi}{3} .
\end{aligned}
$$

At this point, both sets of dihedral angles being greater than $\frac{\pi}{3}$ does not hold, thus A is a sufficient condition for B but not necessary. Hence, the answer is (A).