In a regular triangular pyramid ABCP with vertex P, the side of the base is equal to 2. A section is made through the base side BC, intersecting the edge PA at point M, such that PM:MA = 1:3, and the area of the section is 3. Find the height of the pyramid.
Let $Q$ be the center of the base of the given regular pyramid, $L$ be the midpoint of edge $BC$, and $F$ be the projection of point $M$ onto the base plane. The segment $AQ$ is the orthogonal projection of segment $AP$ onto the base plane, so point $F$ lies on segment $AQ$. Since $MF \parallel PQ$ (as perpendiculars to the same plane $ABC$), we have

$$
\frac{QF}{FA} = \frac{P'M}{MA} = \frac{1}{3}.
$$

From the equality of isosceles triangles $APB$ and $APS$, it follows that segments $BM$ and $CM$ are equal. The median $ML$ of isosceles triangle $BMC$ is its height, so

$$
\frac{1}{2} BC \cdot ML = 3, \quad \text{from which} \quad ML = \frac{6}{BC} = \frac{6}{2} = 3.
$$

Next, we find:

$$
\begin{gathered}
AL = BC \cdot \frac{\sqrt{3}}{2} = \sqrt{3}, \quad LQ = \frac{1}{3} AL = \frac{\sqrt{3}}{3}, \quad AQ = \frac{2}{3} AL = \frac{2 \sqrt{3}}{3} \\
QF = \frac{1}{4} AQ = \frac{\sqrt{3}}{6}, \quad FL = LQ + QF = \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{6} = \frac{\sqrt{3}}{2} \\
MF = \sqrt{ML^2 - FL^2} = \sqrt{9 - \frac{3}{4}} = \frac{\sqrt{33}}{2}
\end{gathered}
$$

From the similarity of triangles $APQ$ and $AMF$, it follows that

$$
PQ = MF \cdot \frac{AP'}{AM} = \frac{4}{3} MF = \frac{4}{3} \cdot \frac{\sqrt{33}}{2} = 2 \sqrt{\frac{11}{3}}
$$

## Answer

$\sqrt{\frac{11}{3}}$