6. For a regular tetrahedron $P-ABC$ with all edge lengths equal to $1$, $L, M, N$ are the midpoints of edges $PA, PB, PC$ respectively. The area of the section formed by the plane $LMN$ cutting through the circumsphere of the tetrahedron is $\qquad$
The height of a regular tetrahedron is \( h = \sqrt{1 - \left(\frac{\sqrt{3}}{3}\right)^2} = \frac{\sqrt{6}}{3} \), the radius of the circumscribed sphere is \( R = \frac{3}{4} h = \frac{\sqrt{6}}{4} \), and the distance from the center of the circumscribed sphere \( O \) to the midsection plane \( L M N \) is \( d = \frac{1}{4} h = \frac{\sqrt{6}}{12} \). Therefore, the radius of the cross-sectional circle is \( r^2 = R^2 - d^2 = \left(\frac{\sqrt{6}}{4}\right)^2 - \left(\frac{\sqrt{6}}{12}\right)^2 = \left(\frac{\sqrt{6}}{4}\right)^2 \left(1 - \frac{1}{9}\right) = \frac{6}{16} \times \frac{8}{9} = \frac{1}{3} \).
Thus, the area of the cross-section of the circumscribed sphere of the tetrahedron by the plane \( L M N \) is \( \frac{\pi}{3} \).