Answer: \(\frac{38}{3}\) square units.

The curve is

\(y=(9-x)(x-3).\)

Expanding,

\(y=-x^2+12x-27.\)

The maximum point occurs on the axis of symmetry, halfway between the roots \(x=3\) and \(x=9\). Hence

\(x=6.\)

So \(A=(6,0)\).

The right-hand intercept is

\(B=(9,0).\)

Since \(C=(k,k-3)\) lies on the curve,

\(k-3=(9-k)(k-3).\)

Because \(k\gt 3\), \(k-3\neq0\), so divide by \(k-3\):

\(1=9-k.\)

Therefore

\(k=8.\)

So

\(C=(8,5).\)

The shaded region consists of a rectangle from \(x=6\) to \(x=8\), height \(5\), plus the area under the curve from \(x=8\) to \(x=9\).

The rectangle area is

\((8-6)\cdot5=10.\)

The curve is

\(y=-x^2+12x-27.\)

So the curved part has area

\(\int_8^9(-x^2+12x-27)\,dx.\)

Integrating,

\(\int(-x^2+12x-27)\,dx =-\frac{x^3}{3}+6x^2-27x.\)

Therefore

\(\int_8^9(-x^2+12x-27)\,dx =\frac83.\)

Hence the total shaded area is

\(10+\frac83=\frac{38}{3}.\)