(a) \[\frac{dy}{dx}=-3x^2+18x-15=-3(x-1)(x-5).\] The stationary points are \((1,-10)\), \((5,22)\). The line through them has gradient \((22+10)/(5-1)=8\), so \[y+10=8(x-1)\quad\Longrightarrow\quad y=8x-18.\]

(b) At \(x=3\), the curve gives \(-27+81-45-3=6\) and the line gives \(24-18=6\); both meet at \((3,6)\).

(c) The difference is \[-x^3+9x^2-15x-3-(8x-18)=-(x-1)(x-3)(x-5).\] The shaded right-hand region runs from \(x=3\) to \(x=5\), where the curve is above the line. Therefore \[\begin{aligned}\text{area}&=\int_3^5(-x^3+9x^2-23x+15)\,dx\\&=\left[-\frac{x^4}{4}+3x^3-\frac{23x^2}{2}+15x\right]_3^5\\&=\frac{25}{4}-\frac94=\boxed{4}.\end{aligned}\]