260. a) In the pyramid $S A B C$, the base is a triangle $A B C$, where $\widehat{B A} C=A, \widehat{C B A}=B$, and the radius of the circumscribed circle around it is $R$. The edge $S C$ is perpendicular to the plane $A B C$. Find $| S C |$, if it is known that $\frac{1}{\sin \alpha}+\frac{1}{\sin \beta}-\frac{1}{\sin \gamma}=1$, where $\alpha, \beta$ and $\gamma$ are the angles formed by the edges $S A, S B$ and $S C$ with the planes of the faces $S B C, S A C$ and $S A B$ respectively.

b) Let $\alpha, \beta$ and $\gamma$ be the angles formed by the edges of a trihedral angle with the planes of the opposite faces.

Prove that $\frac{1}{\sin \alpha}+\frac{1}{\sin \beta}-\frac{1}{\sin \gamma} \geqslant 1$.
260. a) Let $|S C|=d, a, b, c$ be the sides of $\triangle A B C, h_{a}, h_{b}, h_{c}$ the altitudes of $\triangle A B C, s$ its area. Then

$$
\sin \alpha=\frac{h_{a}}{\sqrt{a^{2}+b^{2}}}, \quad \sin \beta=\frac{h_{b}}{\sqrt{d^{2}+a^{2}}}, \quad \sin \gamma=\frac{h_{c}}{\sqrt{d^{2}+h_{c}^{2}}}
$$

Thus, we obtain the equation for $d$

$$
\frac{\sqrt{d^{2}+b^{2}}}{h_{a}}+\frac{\sqrt{d^{2}+a^{2}}}{h_{b}}=1+\frac{\sqrt{d^{2}+h_{c}^{2}}}{h_{\mathbf{c}}}
$$

Multiplying this equation by $2 s$, we get

$$
a \sqrt{d^{2}+b^{2}}+b \sqrt{d^{2}+a^{2}}=2 s+\sqrt{c^{2} d^{2}+4 s^{2}}
$$

Multiplying and dividing both sides of (1) by the differences of the corresponding quantities (assuming $\widehat{A} \neq \widehat{B}$), we get

$$
\frac{a^{2}-b^{2}}{a \sqrt{d^{2}+b^{2}}-b \sqrt{d^{2}+a^{2}}}=\frac{c^{2}}{\sqrt{c^{2} d^{2}+4 s^{2}}-2 s}
$$

from which

$$
a c^{2} \boldsymbol{V} \overline{d^{2}+b^{2}}-b c^{2} \boldsymbol{V} \overline{d^{2}+a^{2}}=\left(a^{2}-b^{2}\right)\left(\sqrt{c^{2} d^{2}+4 s^{2}}-2 s\right)
$$

Multiplying (1) by $b^{2}-a^{2}$ and adding to (2), we get

$a\left(b^{2}+c^{2}-a^{2}\right) \sqrt{a^{2}+b^{2}}+b\left(b^{2}-a^{2}-c^{2}\right) \sqrt{d^{2}+a^{2}}=4 s\left(b^{2}-a^{2}\right)$.

Using the cosine and sine theorems, the last equation can be transformed into

$$
\cos A \cdot \sqrt{d^{2}+b^{2}}-\cos B \cdot \sqrt{d^{2}+a^{2}}=\frac{b^{2}-a^{2}}{2 R}
$$

The right-hand side of equation (3) can be transformed as follows:

$$
\frac{b^{2}-a^{2}}{2 R}=2 R\left(\sin ^{2} B-\sin ^{2} A\right)=2 R \sin (A+B) \sin (B-A)
$$

after which we multiply both sides of (3) by $\cos A \cdot \sqrt{a^{2}+b^{2}}+\cos B \times$ $\times \sqrt{d^{2}+a^{2}}$, we get the equation

$$
\begin{aligned}
& \left(\cos ^{2} A-\cos ^{2} B\right) d^{2}+b^{2} \cos ^{2} A-a^{2} \cos ^{2} B= \\
& =2 R \sin (A+B) \sin (B-A)\left(\cos A \cdot \sqrt{d^{2}+b^{2}}+\cos B \cdot \sqrt{d^{2}+a^{2}}\right)
\end{aligned}
$$

In equation (4), $\cos ^{2} A-\cos ^{2} B=\sin (A+B) \operatorname{sid}(B-A)$, $3^{2} \cos ^{2} A-a^{2} \cos ^{2} B=4 R^{2} \sin (B+A) \sin (B-A)$. Therefore, after simplification, equation (4) transforms into

$$
\cos A \cdot \sqrt{d^{2}+b^{2}}+\cos B \cdot \sqrt{d^{2}+a^{2}}=\frac{d^{2}}{2 R}+2 R
$$

Adding (3) and ( $4^{\prime}$ ), we get

$$
2 \cos A \cdot \sqrt{d^{2}+b^{2}}=\frac{d^{2}}{2 R}+2 R\left(\sin ^{2} B+\cos ^{2} A\right)
$$

from which

$$
\begin{gathered}
\left(\sqrt{d^{2}+b^{2}}-2 R \cos A\right)^{2}=0 \\
d^{2}=4 R^{2}\left(\cos ^{2} A-\sin ^{2} B\right)=4 R^{2} \cos (A+B) \cos (A-B)
\end{gathered}
$$

Thus,

$$
|S C|=2 R \sqrt{\cos (A+B) \cos (A-B)}
$$

The problem has a solution if $A+B<90^{\circ}$, i.e., in triangle $A B C$ angle $C$ is obtuse.

b) Using the notations from part a). Then our inequality can be rewritten as

$$
\frac{\sqrt{d^{2}+a^{2}}}{h_{b}}+\frac{\sqrt{d^{2}+b^{2}}}{h_{a}}-\frac{\sqrt{d^{2}+h_{c}^{2}}}{h_{c}} \geqslant 1
$$

If angle $C$ is acute, then the right-hand side, as follows from part a), is never equal to 1, so the inequality holds since it is satisfied at $d=0$. If, however, $C$ is an obtuse (or right) angle, the right-hand side equals 1 at a unique value of $d$ (if $C$ is a right angle, then $d=0$). But at $d=0$ and sufficiently large values of $d$, the inequality is obvious (for large $d$ it follows from the triangle inequality), so if at some value of $d$ the left-hand side were less than 1, then the left-hand side would take the value 1 at two different values of $d$.