(i) To find \(k\), use the point \(A(2, 2)\) on the curve:

\(2 = k(2^3 - 7 \times 2^2 + 12 \times 2)\)

\(2 = k(8 - 28 + 24)\)

\(2 = k \times 4\)

\(k = \frac{1}{2}\)

(ii) Verify the curve meets \(y = x\) at \(x = 5\):

Substitute \(x = 5\) into the curve equation:

\(y = \frac{1}{2}(5^3 - 7 \times 5^2 + 12 \times 5)\)

\(y = \frac{1}{2}(125 - 175 + 60)\)

\(y = \frac{1}{2}(10) = 5\)

Since \(y = 5\) when \(x = 5\), the curve meets the line \(y = x\).

(iii) Find the area of the shaded region:

The area is given by the integral:

\(\int_0^2 \left( \frac{1}{2}(x^3 - 7x^2 + 12x) - x \right) \, dx\)

Integrate term by term:

\(\int_0^2 \left( \frac{1}{2}x^3 - \frac{7}{2}x^2 + 5x \right) \, dx\)

\(= \left[ \frac{1}{8}x^4 - \frac{7}{6}x^3 + \frac{5}{2}x^2 \right]_0^2\)

Evaluate the definite integral:

\(= \left( \frac{1}{8}(16) - \frac{7}{6}(8) + \frac{5}{2}(4) \right) - 0\)

\(= 2 - \frac{28}{3} + 10\)

\(= \frac{8}{3}\)