(i) To find the coordinates of \(B\), set \(x = 0\) in the equation \(y = \sqrt{1 + 2x}\):

\(y = \sqrt{1 + 2(0)} = \sqrt{1} = 1\)

So, \(B = (0, 1)\).

For \(C\), given \(y = 3\):

\(3 = \sqrt{1 + 2x}\)

\(9 = 1 + 2x\)

\(2x = 8\)

\(x = 4\)

So, \(C = (4, 3)\).

(ii) The derivative \(\frac{dy}{dx}\) is:

\(\frac{dy}{dx} = \frac{1}{2} \times 2(1 + 2x)^{-\frac{1}{2}} = \frac{1}{\sqrt{1 + 2x}}\)

At \(C(4, 3)\):

\(\frac{dy}{dx} = \frac{1}{\sqrt{9}} = \frac{1}{3}\)

The gradient of the normal is \(-3\).

Equation of the normal:

\(y - 3 = -3(x - 4)\)

or \(y = -3x + 15\).

(iii) To find the volume, use the formula for the volume of revolution:

\(y^2 = 1 + 2x \Rightarrow x = \frac{1}{2}(y^2 - 1)\)

Volume \(V = \pi \int_0^1 x^2 \, dy\)

\(V = \pi \int_0^1 \left( \frac{1}{2}(y^2 - 1) \right)^2 \, dy\)

\(V = \pi \times \frac{1}{4} \int_0^1 (y^4 - 2y^2 + 1) \, dy\)

\(V = \pi \times \frac{1}{4} \left[ \frac{y^5}{5} - \frac{2y^3}{3} + y \right]_0^1\)

\(V = \pi \times \frac{1}{4} \left[ \frac{1}{5} - \frac{2}{3} + 1 \right]\)

\(V = \frac{2}{15} \pi\)