Parabola

It is noted that when $x \to \infty$ , $f(x) \to \infty$ . We need to find the values of $p$ such that $\min (f(x)) = 8$ for $x \in [2,\infty)$ . To do that we can find the value of $p$ , where $f'(x) = 0$ if $f''(x) > 0$ .

$\begin{aligned} f'(x) & = 2x - (p-2) & \small \color{#3D99F6} \text{For }f'(x) = 0 \\ \implies x & = \frac {p - 2}2 & \small \color{#3D99F6} \text{Note that } f''(x) = 2 > 0 \\ \implies \min (f(x)) & = f\left(\frac {p - 2}2\right) = 8\end{aligned}$

Therefore,

$\begin{aligned} \left(\frac {p - 2}2\right)^2 - (p-2)\left(\frac {p - 2}2\right)+3p-2 & = 8 \\ p^2 - 16p + 44 & = 0 \\ \implies p & = 8 + 2\sqrt 5 & \small \color{#3D99F6} \text{For }p=8-2\sqrt 5, \ \frac {p-2}2 < 2 \text{ rejected.} \end{aligned}$

For a strictly increasing $f$ , $\min (f(x)) = f(2) = 8$ . Then:

$\begin{aligned} 2^2 - 2(p-2) + 3p - 2 & = 8 \\ \implies p & = 2 \end{aligned}$

$\implies f(x) = x^2 + 4$ , a strictly increasing function.

Therefore, the sum of possible values of $p$ is $8+2\sqrt5+2=10+2\sqrt5$ , $\implies \dfrac a{bc} = \dfrac {10}{2\times 5} = \boxed{1}$