2.3. Find the volume of a regular triangular pyramid, the height of which is equal to $h$, and the planar angles at the vertex are right angles.
2.3. Let's draw $B D \perp A C$ (Fig. 2.3). Denote the side of the base by $a$. Since the pyramid is regular, $O D=\frac{a \sqrt{3}}{6}$, $A D=D C=\frac{a}{2}$, and $S_{\mathrm{och}}=\frac{a^{2} \sqrt{3}}{4}$. Express $a$ in terms of $h$. In $\triangle S A C$ we have

$S A=S C, \angle C S A=90^{\circ}$, hence $\angle A S D=45^{\circ}$ and $S D=A D=\frac{a}{2}$. Then

$$
h=S O=\sqrt{S D^{2}-O D^{2}}=\sqrt{\frac{a^{2}}{4}-\frac{3 a^{2}}{36}}=\frac{a}{\sqrt{6}}
$$

and, therefore, $a=h \sqrt{6}$. Thus,

$$
\begin{aligned}
& \qquad V=\frac{1}{3} S_{\text {ocн }} h=\frac{1}{3} \cdot \frac{a^{2} \sqrt{3}}{4} \cdot h=\frac{1}{3} \cdot \frac{6 h^{2} \sqrt{3}}{4} \cdot h=\frac{h^{3} \sqrt{3}}{2} \\
& \text { Answer: } \frac{h^{3} \sqrt{3}}{2}
\end{aligned}
$$