Problem 996

320. Suppose the opposite. Let the planes in which the arcs are located intersect pairwise on the surface of the sphere at points $A$ and $A_{1}, B$ and $B_{1}, C$ and $C_{1}$ (Fig. 65). Since each arc is greater than $180^{\circ}$, it must contain at least one of any two opposite points of the circle on which it is located.

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Fig. 65. Let us label these arcs according to the planes in which they are located as I, II, III. $A$ and $A_{1}$ are the points of intersection of planes I and II, $B$ and $B_{1}$ are the points of intersection of planes II and III, and $C$ and $C_{1}$ are the points of intersection of planes III and I. Each of the points $A, A_{1}, B, B_{1}, C, C_{1}$ must belong to one arc. Suppose $A_{1}$ and $C_{1}$ belong to arc I, $B_{1}$ belongs to arc II. Then $B$ and $C$ must belong to arc III, and $A$ must belong to arc II. Let $\alpha, \beta, \gamma$ be the plane angles of the trihedral angles as shown in the figure, and $O$ be the center of the sphere. Since arc I does not contain points $A$ and $C$, the inequality $360^{\circ} - \beta > 300^{\circ}$ must hold.

Similarly, since arc II does not contain points $B$ and $A_{1}$, it must be $180^{\circ} + \alpha > 300^{\circ}$, and finally, for arc III, we have $360^{\circ} - \gamma > 300^{\circ}$. Thus, $\beta < 60^{\circ}$, $\alpha < 120^{\circ}$, and $\gamma < 60^{\circ}$. Therefore, $\alpha + \beta + \gamma < 240^{\circ}$, which is impossible.