(i) To find the equation of the normal at \(A\), first find the derivative of the curve \(y = 2 - \frac{18}{2x+3}\).

\(\frac{dy}{dx} = 18(2x+3)^{-2} \times 2\).

At \(x = 3\), the slope \(m\) of the tangent is \(\frac{4}{9}\).

The slope of the normal is \(-\frac{9}{4}\) (since \(m_1 m_2 = -1\)).

The equation of the normal is \(y = -\frac{9}{4}(x-3)\).

Rearranging gives \(4y + 9x = 27\).

(ii) The normal meets the y-axis at \((0, 6\frac{3}{4})\).

The curve meets the y-axis at \((0, -4)\).

Thus, the length of \(BC\) is \(10\frac{3}{4}\).