The roots of the equation $\sqrt{2}x^2 - \frac{3}{\sqrt{2}}x + c = 0$ are $p$ and $2p$.

Let $a > 0$, and one root of the equation $a^2x^2 + 12a - 7 = 0$ is $64(p^6 + c^{12})$.

What is the value of $a$?

The equation $\sqrt{2}x^2 - \frac{3}{\sqrt{2}}x + c = 0$ has roots $p$ and $2p$.

Using Vieta's formulas (relationships between roots and coefficients):

Relationship	Formula	Substitution
Sum of roots	$p + 2p = \frac{3/\sqrt{2}}{\sqrt{2}}$	$3p = \frac{3}{2}$
Product of roots	$p \cdot 2p = \frac{c}{\sqrt{2}}$	$2p^2 = \frac{c}{\sqrt{2}}$

From the sum: $3p = \frac{3}{2}$ gives us $p = \frac{1}{2}$

From the product: $2p^2 = \frac{c}{\sqrt{2}}$

Substituting $p = \frac{1}{2}$:

$2 \cdot \left(\frac{1}{2}\right)^2 = \frac{c}{\sqrt{2}}$

$2 \cdot \frac{1}{4} = \frac{c}{\sqrt{2}}$

$\frac{1}{2} = \frac{c}{\sqrt{2}}$

$c = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$

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Now we calculate $64(p^6 + c^{12})$:

$p^6 = \left(\frac{1}{2}\right)^6 = \frac{1}{64}$

$c^{12} = \left(\frac{1}{\sqrt{2}}\right)^{12} = \frac{1}{(\sqrt{2})^{12}} = \frac{1}{(2^{1/2})^{12}} = \frac{1}{2^6} = \frac{1}{64}$

Therefore: $64(p^6 + c^{12}) = 64\left(\frac{1}{64} + \frac{1}{64}\right) = 64 \cdot \frac{2}{64} = 2$

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So one root of the equation $a^2x^2 + 12a - 7 = 0$ is $x = 2$.

Substituting $x = 2$ into the equation:

$a^2(2)^2 + 12a - 7 = 0$

$4a^2 + 12a - 7 = 0$

Using the quadratic formula where $A = 4$, $B = 12$, $C = -7$:

$a = \frac{-12 \pm \sqrt{144 + 112}}{8} = \frac{-12 \pm \sqrt{256}}{8} = \frac{-12 \pm 16}{8}$

This gives: $a = \frac{4}{8} = \frac{1}{2}$ or $a = \frac{-28}{8} = -\frac{7}{2}$

Since $a > 0$, we have $a = \frac{1}{2}$

Answer: Option 2