The diagram shows part of the curve $y=(9-x)(x-3)$ and the line $y=k-3$, where $k>3$. The line through the maximum point of the curve, parallel to the y-axis, meets the x-axis at A. The curve meets the x-axis at B and the line $y=k-3$ meets the curve at the point $C(k,\,k-3)$.

1. Substitute the coordinates of point $C$ into the equation of the curve.

Since $C(k, k-3)$ lies on the curve $y = (9-x)(x-3)$, we substitute $x = k$ and $y = k-3$:
$k - 3 = (9 - k)(k - 3)$

2. Solve the equation for $k$.

$k - 3 - (9 - k)(k - 3) = 0$
$(k - 3)[1 - (9 - k)] = 0$
$(k - 3)(k - 8) = 0$
$k = 3 \quad \text{or} \quad k = 8$

3. Select the valid value of $k$ based on the given condition.

Since we are given $k > 3$, we reject $k = 3$.
Therefore, $k = 8$ (shown).

1. Find the x-coordinate of the maximum point A

The curve is $y = (9-x)(x-3)$.
The x-intercepts are $x = 9$ and $x = 3$.
By symmetry, the x-coordinate of the maximum point A is $\frac{9+3}{2} = 6$.

2. Find the x-coordinate of point B

The curve meets the x-axis at B, which is the right-hand intercept.
From the roots $x=3$ and $x=9$, B is at $x=9$.

3. Set up the definite integral for the shaded area

The shaded region is bounded by the curve, the x-axis, and the vertical line through A ($x=6$).
Area = $\int_{6}^{9} (9-x)(x-3) \, dx$

4. Expand the integrand and integrate

$(9-x)(x-3) = 9x - 27 - x^2 + 3x = -x^2 + 12x - 27$
$\int (-x^2 + 12x - 27) \, dx = \left[ -\frac{x^3}{3} + 6x^2 - 27x \right]_{6}^{9}$

5. Evaluate the definite integral

At $x=9$:
$-\frac{9^3}{3} + 6(9^2) - 27(9) = -243 + 486 - 243 = 0$

At $x=6$:
$-\frac{6^3}{3} + 6(6^2) - 27(6) = -72 + 216 - 162 = -18$

Area = $0 - (-18) = 18$