(i) Midpoint of [AC] with \( A(6, 0) \), \( C(3, 3\sqrt{3}) \):

\[ M = \left( \frac{6 + 3}{2}, \frac{0 + 3\sqrt{3}}{2} \right) = \left( \frac{9}{2}, \frac{3\sqrt{3}}{2} \right) = \left( 4.5, \frac{3\sqrt{3}}{2} \right) \]

Answer: \( \left( 4.5, \frac{3\sqrt{3}}{2} \right) \)

(ii) Since OABC is symmetrical about [OB], \( M \) lies on [OB]. Line [OB] passes through \( O(0, 0) \) and \( M \left( 4.5, \frac{3\sqrt{3}}{2} \right) \).

Slope of [OB]:

\[ m = \frac{\frac{3\sqrt{3}}{2} – 0}{4.5 – 0} = \frac{\frac{3\sqrt{3}}{2}}{4.5} = \frac{3\sqrt{3}}{2} \cdot \frac{1}{4.5} = \frac{3\sqrt{3}}{9} = \frac{\sqrt{3}}{3} \]

Equation through origin: \( y = \frac{\sqrt{3}}{3}x \).

Answer: \( y = \frac{\sqrt{3}}{3}x \)

Since [OA] \(\perp\) [AB], find coordinates of \( B \). Let \( B(x_B, y_B) \) lie on \( y = \frac{\sqrt{3}}{3}x \), so \( y_B = \frac{\sqrt{3}}{3}x_B \).

Slope of [OA] (from \( O(0, 0) \) to \( A(6, 0) \)): \( m_{OA} = \frac{0 – 0}{6 – 0} = 0 \).

Slope of [AB] (from \( A(6, 0) \) to \( B(x_B, y_B) \)): \( m_{AB} = \frac{y_B – 0}{x_B – 6} = \frac{y_B}{x_B – 6} \).

Since \( m_{OA} \cdot m_{AB} = -1 \), and \( m_{OA} = 0 \), \( m_{AB} \) is undefined (vertical line). Thus, \( x_B = 6 \).

Substitute \( x_B = 6 \) into \( y_B = \frac{\sqrt{3}}{3}x_B \):

\[ y_B = \frac{\sqrt{3}}{3} \cdot 6 = 2\sqrt{3} \]

So, \( B(6, 2\sqrt{3}) \).

Verify symmetry about [OB]: Midpoint of [AC] matches \( M \). For [BC], midpoint:

\[ M_{BC} = \left( \frac{6 + 3}{2}, \frac{2\sqrt{3} + 3\sqrt{3}}{2} \right) = \left( 4.5, \frac{5\sqrt{3}}{2} \right) \]

This does not lie on [OB], indicating a need to check coordinates or symmetry. Assume \( C’ \), the symmetric point to \( C \), exists. Since \( C(3, 3\sqrt{3}) \), find \( C’ \) such that midpoint of [CC’] is on [OB]. Let \( C'(x_C’, y_C’) \), midpoint:

\[ \left( \frac{3 + x_C’}{2}, \frac{3\sqrt{3} + y_C’}{2} \right) \]

Lies on \( y = \frac{\sqrt{3}}{3}x \):

\[ \frac{3\sqrt{3} + y_C’}{2} = \frac{\sqrt{3}}{3} \cdot \frac{3 + x_C’}{2} \]

Assume \( x_C’ = 6 \), then midpoint \( x = 4.5 \), \( y = \frac{\sqrt{3}}{3} \cdot 4.5 = \frac{3\sqrt{3}}{2} \), matches \( M \).

Area of triangle OAB: \( \text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 6 \cdot 2\sqrt{3} = 6\sqrt{3} \).

By symmetry, area of OABC = \( 2 \cdot 6\sqrt{3} = 12\sqrt{3} \).

Answer: \( 12\sqrt{3} \)