What about this? -1

Using Vieta's formula: $\displaystyle\sum_{\text{cyc}}\alpha=7,\displaystyle\sum_{\text{cyc}}\alpha\beta=5,\alpha\beta\gamma=-2$

$\mathcal A=6\displaystyle\sum_{\text{cyc}}\left(\dfrac{\alpha^2}{\underbrace{\alpha\beta\gamma}_{-2}+\alpha}\right)$

$= 6\displaystyle\sum_{\text{cyc}}\left(\dfrac{\overbrace{\alpha^2}^{ \color{#D61F06}{4}+(\alpha^2-\color{#D61F06}{4})}}{\alpha-2}\right)$

$=6\displaystyle\sum_{\text{cyc}}\left(\dfrac{4}{\alpha-2}+\alpha+2\right)$

Now $\small{\displaystyle\sum_{\text{cyc}}\dfrac{1}{\alpha-2}}$ can be found using $\large\color{#EC7300}{3~methods}$ :

---

$(1).$ Take LCM so that it finally comes out to be $\small{\dfrac{\displaystyle\sum_{\text{cyc}}\alpha\beta-4\displaystyle\sum_{\text{cyc}}\alpha+12}{4\displaystyle\sum_{\text{cyc}}\alpha-2\displaystyle\sum_{\text{cyc}}\alpha\beta+ \alpha\beta\gamma-8}}$ and just put values.

$(2).$ Transform the polynomial by putting $x=\dfrac{2y+1}{y}$ and find its sum of roots using Vieta's formula.

$(3).$ Write $f(x)=(x-\alpha)(x-\beta)(x-\gamma$ ) , take $\ln$ and differentiate both sides to get : $\small{\dfrac{-f'(x)}{f(x)}=\displaystyle\sum_{\text{cyc}}\dfrac1{\alpha-x}}$ and just put $x=2$ so that finally you'll get $\displaystyle\sum_{\text{cyc}}\dfrac{1}{\alpha-2}=\dfrac{-11}{8}$ , so that:

---

$\mathcal A=6\left(4\times\dfrac{-11}8+7+6\right)=\boxed{45}$

So the answer is $\color{#3D99F6}{\boxed{\text{None of the other options}}}$ .

$\begin{aligned} \mathcal A & = \sum_\text{cyc} \frac {6\alpha}{\beta \gamma +1} \\ & = 6 \sum_\text{cyc} \frac \alpha {\frac {\color{#3D99F6}{\alpha \beta \gamma}} {\alpha} +1} \quad \quad \small \color{#3D99F6}{\text{Vieta's formula: }\alpha \beta \gamma = -2 } \\ & = 6 \sum_\text{cyc} \frac {\alpha^2} {\alpha -2} \\ & = 6 \sum_\text{cyc} \frac {\alpha^2-4+4} {\alpha -2} \\ & = 6 \sum_\text{cyc} \frac {(\alpha-2)(\alpha+2)+4} {\alpha -2} \\ & = 6 \sum_\text{cyc} \left(\alpha+2 + \frac 4 {\alpha -2}\right) \\ & = 6\left(\color{#3D99F6}{\alpha + \beta + \gamma} + 6 + 4\left(\frac 1 {\alpha -2} + \frac 1 {\beta -2} + \frac 1 {\gamma -2} \right) \right) \quad \quad \small \color{#3D99F6}{\text{Vieta's formula: }\alpha+\beta+\gamma = 7} \\ & = 6\left(\color{#3D99F6}{7} + 6 + 4\left(\frac {\color{#3D99F6}{\alpha \beta + \beta \gamma + \gamma \alpha} -4(\color{#3D99F6}{\alpha + \beta + \gamma})+12}{-\color{#D61F06}{(2-\alpha)(2-\beta)(2-\gamma)}} \right) \right) \quad \quad \small \color{#3D99F6}{\text{Vieta's: }\alpha \beta + \beta \gamma + \gamma \alpha = 5} \\ & = 78 + 24\left(\frac {\color{#3D99F6}{5} -4(\color{#3D99F6}{7})+12}{-\color{#D61F06}{(-8)}} \right) \quad \quad \small \color{#D61F06}{\text{As }(2-\alpha)(2-\beta)(2-\gamma) = 2^3 -7(2^2) +5(2)+2 = -8} \\ & = 45 \quad \boxed{\text{None of the other options}} \end{aligned}$

A solution that uses Newton's sums:

Note that for my solution, let $\alpha = a,\;\beta=b,\;\gamma=c$

From Vieta's formula, we get

$\color{#3D99F6}{a+b+c = 7}\\ \color{#D61F06}{ab+ac+bc = 5}\\ \color{#EC7300}{abc = -2}$

$\mathcal A = \displaystyle \sum_{\text{cyc}} \dfrac{6a}{bc+1}\\ =\displaystyle \sum_{\text{cyc}} \dfrac{6a^2}{\color{#EC7300}{abc}+a}\\ =\displaystyle \sum_{\text{cyc}} \dfrac{6a^2}{a-2}\\ =\dfrac{6a^2}{a-2}+\dfrac{6b^2}{b-2}+\dfrac{6c^2}{c-2}\\ =\dfrac{6a^2(b-2)(c-2) + 6b^2(a-2)(c-2)+6c^2(a-2)(b-2)}{(a-2)(b-2)(c-2)}\\ =\dfrac{6a^2\big(\color{#EC7300}{bc} - 2(\color{#3D99F6}{b+c})+4\big)+6b^2\big(\color{#EC7300}{ac} - 2(\color{#3D99F6}{a+c})+4\big)+6c^2\big(\color{#EC7300}{ab} - 2(\color{#3D99F6}{a+b})+4\big)}{\color{#EC7300}{abc} -2(\color{#D61F06}{ab+ac+bc})+4(\color{#3D99F6}{a+b+c}) -8}\\ =\dfrac{6a^2\big(\color{#EC7300}{\frac{-2}{a}} - 2(\color{#3D99F6}{7-a})+4\big)+6b^2\big(\color{#EC7300}{\frac{-2}{b}} - 2(\color{#3D99F6}{7-b})+4\big)+6c^2\big(\color{#EC7300}{\frac{-2}{c}} - 2(\color{#3D99F6}{7-c})+4\big)}{\color{#EC7300}{-2} -2(\color{#D61F06}{5})+4(\color{#3D99F6}{7}) -8}\\ =\dfrac{6a^2(\frac{-2}{a}-10+2a)+6b^2(\frac{-2}{b}-10+2b)+6c^2(\frac{-2}{c}-10+2c)}{-2-10+28 -8}\\ =\dfrac{-12a-60a^2+12a^3-12b-60b^2+12b^3-12c-60c^2+12c^3}{8}\\ =\dfrac{12(\color{#69047E}{a^3+b^3+c^3})-60(\color{#20A900}{a^2+b^2+c^2})-12(\color{#3D99F6}{a+b+c})}{8}$

Now, we apply Newton's sums:

$\color{#3D99F6}{a+b+c = 7}\\ \color{#20A900}{a^2+b^2+c^2} = (\color{#3D99F6}{a+b+c})^2 - 2(\color{#D61F06}{ab+ac+bc}) = (\color{#3D99F6}{7})^2 - 2(\color{#D61F06}{5}) = \color{#20A900}{39}\\ \color{#69047E}{a^3+b^3+c^3} = (\color{#3D99F6}{a+b+c})(\color{#20A900}{a^2+b^2+c^2}) - (\color{#D61F06}{ab+ac+bc})(\color{#3D99F6}{a+b+c}) + 3(\color{#EC7300}{abc}) = (\color{#3D99F6}{7})(\color{#20A900}{39}) - (\color{#D61F06}{5})(\color{#3D99F6}{7}) + 3(\color{#EC7300}{-2}) =\color{#69047E}{232}$

Substitute these values in:

$\mathcal A = \dfrac{12(\color{#69047E}{a^3+b^3+c^3})-60(\color{#20A900}{a^2+b^2+c^2})-12(\color{#3D99F6}{a+b+c})}{8}\\ =\dfrac{12(\color{#69047E}{232})-60(\color{#20A900}{39})-12(\color{#3D99F6}{7})}{8}\\ =\dfrac{2784-2340-84}{8}\\ =\dfrac{360}{8}\\ =45$

Therefore, the answer is $\boxed{\text{None of the other options}}$ (Though I'd nitpick and say that this is grammatically incorrect, it should be "None of the options above" instead)

Took lcm multiplied the brackets and then substituded using vieta's formula was a bit long method but i got 45 as ans