(a.i) Discriminant Value:

For \( f(x) = 3x^2 – 6x + p \), the discriminant is:
\( D = b^2 – 4ac = (-6)^2 – 4(3)(p) = 36 – 12p \)
Since there are two equal roots: \( D = \boxed{0} \)

(a.ii) Showing p = 3:

From part (a.i): \( 36 – 12p = 0 \)
\( \Rightarrow 36 = 12p \)
\( \Rightarrow p = \boxed{3} \) ✓

(b) Vertex Coordinates:

Method 1: Using vertex formula
\( x \)-coordinate: \( x = -\frac{b}{2a} = \frac{6}{6} = \boxed{1} \)
\( y \)-coordinate: \( f(1) = 3(1)^2 – 6(1) + 3 = \boxed{0} \)
Vertex: \( \boxed{(1, 0)} \)

Method 2: Completing the square
\( f(x) = 3(x^2 – 2x) + 3 = 3[(x-1)^2 – 1] + 3 = 3(x-1)^2 \)
Confirms vertex at \( (1, 0) \)

(c) Solution of f(x) = 0:

Since vertex is on x-axis and parabola opens upwards:
\( \boxed{x = 1} \) (double root)

(d) Vertex Form Parameters:

From completed square form \( f(x) = 3(x-1)^2 + 0 \):
(i) \( a = \boxed{3} \) (vertical stretch factor)
(ii) \( h = \boxed{1} \) (x-coordinate of vertex)
(iii) \( k = \boxed{0} \) (y-coordinate of vertex)

(e) Transformation to g(x):

Step 1: Reflect \( f(x) \) in x-axis
\( -f(x) = -3x^2 + 6x – 3 \)

Step 2: Translate 6 units up
\( g(x) = -3x^2 + 6x – 3 + 6 \)
\( g(x) = \boxed{-3x^2 + 6x + 3} \)