11.120. The lateral surface area of a regular triangular pyramid is three times the area of the base. The area of the circle inscribed in the base is numerically equal to the radius of this circle. Find the volume of the pyramid.
Solution.

Let $DO$ be the height of the regular pyramid $DABC$, $K$ be the midpoint of $AB$ (Fig. 11.17). Let the radius of the circle inscribed in $\triangle ABC$, $OK=r$.

Then, by the condition, $\pi r^{2}=r$, from which $r=\frac{1}{\pi}$.

$AB=2 r \sqrt{3}, S_{\triangle ABC}=\frac{AB^{2} \sqrt{3}}{4}=3 r^{2} \sqrt{3}$. The lateral surface area $S_{6}=3 S_{\triangle ADB}$. By the condition $S_{6}=3 S_{\triangle ABC}$, then $S_{\triangle ADB}=S_{\triangle ABC}$;

$\frac{1}{2} AB \cdot DK=3 r^{2} \sqrt{3} ; r \sqrt{3} \cdot DK=3 r^{2} \sqrt{3} ; DK=3 r$.

From $\triangle DOK\left(\angle DOK=90^{\circ}\right): DO=\sqrt{DK^{2}-OK^{2}}=2 r \sqrt{2}$. Volume of the pyramid: $V=\frac{1}{3} \cdot 3 r^{2} \sqrt{3} \cdot 2 r \sqrt{2}=2 r^{3} \sqrt{6}=\frac{2 \sqrt{6}}{\pi^{3}}$.

Answer: $\frac{2 \sqrt{6}}{\pi^{3}}$.