(i) At \(x = 2\), the tangent has gradient \(3\). Therefore, the normal has gradient \(-\frac{1}{3}\) since \(m_1 m_2 = -1\).

The equation of the normal is \(y - 11 = -\frac{1}{3}(x - 2)\).

(ii) To find the equation of the curve, integrate \(\frac{dy}{dx} = \frac{6}{\sqrt{3x - 2}}\).

Integrate: \(y = 6 \int \frac{1}{\sqrt{3x - 2}} \, dx\).

This gives \(y = 4\sqrt{3x - 2} + c\).

Using the point \((2, 11)\) to find \(c\):

\(11 = 4\sqrt{3(2) - 2} + c\)

\(11 = 4\sqrt{4} + c\)

\(11 = 8 + c\)

\(c = 3\)

Thus, the equation of the curve is \(y = 4\sqrt{3x - 2} + 3\).