(a) To find \(BC\), use the cosine rule in triangle \(OBC\):

\(BC^2 = r^2 + (2r)^2 - 2 \cdot r \cdot 2r \cdot \cos\left(\frac{\pi}{6}\right)\)

\(BC^2 = r^2 + 4r^2 - 4r^2 \cdot \frac{\sqrt{3}}{2}\)

\(BC^2 = 5r^2 - 2r^2\sqrt{3}\)

\(BC = r\sqrt{5 - 2\sqrt{3}}\)

(b) The perimeter of the shaded region is the arc length of \(AB\) plus \(AC\) and \(BC\):

Arc length \(AB = \frac{2\pi r}{6}\)

\(AC = r\)

Perimeter = \(\frac{2\pi r}{6} + r + r\sqrt{5 - 2\sqrt{3}}\)

(c) The area of the shaded region is the area of the sector minus the area of triangle \(OAB\):

Sector area = \(\frac{1}{2} \cdot 4r^2 \cdot \frac{\pi}{6}\)

Triangle area = \(\frac{1}{2} \cdot r \cdot 2r \cdot \sin\left(\frac{\pi}{6}\right)\)

Shaded area = \(r^2 \left( \frac{\pi}{3} - \frac{1}{2} \right)\)