What is Phi doing here?

We Have the following cases to study :

Case 1 : The exponent is 0 .

Case 2 : The Expression $(x^2-7x+12)$ is equal to 1.

Case 3 : The expression $(x^2-7x+12)$ is equal to -1 and the exponent is even .

We observe that $(x^2-3x+2) = x(x-1) - 2(x-1)$ and thus it is always even .

Case 1 : $x^2-3x+2 = 0$

       or, (x-1)(x-2) = 0

So $x = 1 or 2$

Case 2 : $x^2-7x+12 = 1$

Solving the quadratic we get two values of x ,

or, $x = 3+\phi$ or $x = 4-\phi$ [ $\phi=\frac{1+\sqrt{5}}{2}$ ]

Case 3 : $x^2-7x+13 = 0$

D = $7^2-4.1.13 < 0$ So the case is ruled out.

We obtain $\alpha=1,\beta=2,\gamma=3,\zeta=4$ .

Their product

= $1.2(3+\phi)(4-\phi)$

= $(-2)(\phi)^2$ + $2\phi$ + 24

Therefore $|a| + |b| + |c| = 28$

Let $f(x) = x^2 - 7x + 12$ and $g(x) = x^2-3x+2$ and for $f(x)^{g(x)} = 1$ there are two cases.

$\begin{cases} \text{Case 1:} & g(x) = 0 \text{ and } f(x) \ne 0 & \Rightarrow x^2-3x+2 = 0 \\ & & \space (x-1)(x-2) = 0 \\ & & \Rightarrow x = \begin{cases} \color{#3D99F6}{1} \text{ and } f(1) \ne 0 \\ \color{#3D99F6}{2} \text{ and } f(2) \ne 0 \end{cases} \\ \text{Case 2:} & f(x) = x^2 - 7x + 12 = 1 & \Rightarrow x^2 - 7x + 11 = 0 \\ & & \Rightarrow x = \begin{cases} \dfrac{7+\sqrt{5}}{2} = \color{#3D99F6}{3 + \varphi} \\ \dfrac{7-\sqrt{5}}{2} = \color{#3D99F6}{4 - \varphi} \end{cases} \end{cases}$

Therefore, the product of root is: $\begin{aligned} (1)(2)(3+\varphi)(4-\varphi) & = 2(12+\varphi - \varphi^2) = -2\varphi^2 + 2\varphi + 24 \end{aligned}$

$\Rightarrow |a|+|b|+|c| = |-2|+|2|+|24| = 2+2+24 = \boxed{28}$

You should specify $a_1\,<\,a_2\,<\,a_3\,<\,a_4$ in the question.