Find $C$

Let $A = \sqrt[4]{\sqrt 3a+\sqrt 2b}$ . Then $A^4 = 9a^4 + 12\sqrt 6a^3b + 36a^2b^2 + 8\sqrt 6ab^3 + 4 = 49 + 20\sqrt 6$ . Equating the rational and irrational parts, we have $9a^4 + 36a^2b^2 + 4 = 49$ and $12a^3b + 8 ab^3 = 20$ , $\implies a=b=1$ , $\implies A = \sqrt 3 + \sqrt 2$ .

Now consider $B$ :

$\begin{aligned} B & = 8\sqrt 3 + \frac {8\sqrt 6}{\sqrt 3} + \frac {16}{\sqrt 3} + \cdots \\ & = 2^3\cdot 3^\frac 12 + 2^\frac 72 \cdot 3^0 + 2^4 \cdot 3^{-\frac 12} + \cdots \\ & = 2^3\cdot 3^\frac 12 \left(1 + 2^\frac 12 \cdot 3^{-\frac 12} + 2 \cdot 3^{-1} + \cdots \right) \\ & = 8\sqrt 3 \sum_{k=0}^\infty \left(\sqrt{\frac 23} \right)^k \\ & = 8\sqrt 3 \left(\frac 1{1-\sqrt{\frac 23}} \right) \\ & = \frac {24}{\sqrt 3-\sqrt 2} \end{aligned}$

Then the quadratic equation becomes

$\begin{aligned} A(\sqrt 3-\sqrt 2)x^2 + \frac B{\sqrt 3+\sqrt 2}x + C & = 0 \\ (\sqrt 3+\sqrt 2)(\sqrt 3-\sqrt 2)x^2 + \frac {24}{(\sqrt 3+\sqrt 2)(\sqrt 3-\sqrt 2)}x + C & = 0 \\ \implies x^2 + 24x + C & = 0 \end{aligned}$

Consider $k$ .

$\begin{aligned} k & = \log_6 10 - 2\log_6 \sqrt 5 + \log_6 \sqrt{\log_6 18 + \log_6 72} \\ & = \log_6 2 + \log_6 5 - \log_6 5 + \frac 12 \log_6 (\log_6 3+\log_6 6+\log_6 2+2\log_6 6) \\ & = \log_6 2 + \frac 12 \log_6 4 \\ & = 2 \log_6 2 \end{aligned}$

Then $|E-F| = (6\sqrt 6)^k = 6^{\frac 32k} = 6^{\frac 32 \times 2 \log_6 2} = 6^{\log_6 8} = 8$ .

Now it is given that $E$ and $F$ are the roots of the quadratic equation. Then by Vieta's formula, we have $E+F=-24$ and $EF=C$ . Therefore,

$\begin{cases} (E+F)^2 = E^2 + 2EF + F^2 = E^2 + 2C + F^2 = 24^2 & \implies E^2 + 2C + F^2 = 576 & ...(1) \\ |E-F|^2 = E^2 - 2EF + F^2 = E^2 - 2C + F^2 = 8^2 & \implies E^2 - 2C + F^2 = 64 & ...(2)\end{cases}$

$\implies (1) - (2): \quad 4C = 512 \implies C = \boxed{128}$