(i) Equate the line and curve equations: \(-2x + k = \frac{2}{x - 3}\).

Multiply through by \(x - 3\) to clear the fraction: \(-2x(x - 3) + k(x - 3) = 2\).

Simplify to get \(2x^2 - (6 + k)x + (2 + 3k) = 0\).

(ii) For the line to be tangent to the curve, the quadratic equation must have a repeated root, so the discriminant \(b^2 - 4ac = 0\).

Here, \(a = 2\), \(b = -(6 + k)\), \(c = 2 + 3k\).

Calculate the discriminant: \((6 + k)^2 - 4(2)(2 + 3k) = 0\).

Simplify: \(k^2 - 12k + 20 = 0\).

Factorize: \((k - 10)(k - 2) = 0\).

Thus, \(k = 2\) or \(k = 10\).

(iii) For \(k = 2\), substitute into the quadratic: \(2(x - 2)^2 = 0\).

\(x = 2\), \(y = -2\), so \(A = (2, -2)\).

For \(k = 10\), substitute into the quadratic: \(2(x - 4)^2 = 0\).

\(x = 4\), \(y = 2\), so \(B = (4, 2)\).

Find the equation of line \(AB\):

Slope \(m = \frac{2 - (-2)}{4 - 2} = 2\).

Using point-slope form: \(y + 2 = 2(x - 2)\) or \(y = 2x - 6\).