(i) To find the points of intersection, set the equations equal: \(x^2 - x + 3 = 3x + a\). Rearrange to get \(x^2 - 4x + (3 - a) = 0\).

(ii) Substitute \(x = -1\) into the equation \(x^2 - 4x + (3 - a) = 0\):

\((-1)^2 - 4(-1) + (3 - a) = 0\)

\(1 + 4 + 3 - a = 0\)

\(8 - a = 0 \Rightarrow a = 8\)

Substitute \(a = 8\) back into the equation:

\(x^2 - 4x - 5 = 0\)

Factorize: \((x - 5)(x + 1) = 0\)

The other \(x\)-coordinate is \(x = 5\).

(iii) For the line to be tangent, the discriminant must be zero:

\(b^2 - 4ac = 0\)

\((-4)^2 - 4(1)(3 - a) = 0\)

\(16 - 4(3 - a) = 0\)

\(16 - 12 + 4a = 0\)

\(4a = -4 \Rightarrow a = -1\)

Substitute \(a = -1\) into \(x^2 - 4x + 4 = 0\):

\((x - 2)^2 = 0 \Rightarrow x = 2\)

Substitute \(x = 2\) into \(y = x^2 - x + 3\):

\(y = 2^2 - 2 + 3 = 5\)

The coordinates of \(P\) are \((2, 5)\).