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 equation $$a^{2}x^{2} + 12a - 7 = 0$$ is $$64\left(p^{6}+c^{12}\right)$$.
What is the value of a ?

Given, $$\sqrt{\ 2}x^2\ -\ \frac{\ 3}{\sqrt{\ 2}}x\ +c=0$$

Multiplying the equation by $$\sqrt{\ 2}$$ we get,

$$2x^2-3x+\sqrt{\ 2}c=0$$

Roots to this equation are p and 2p.

We know that the sum of roots of a quadratic equation $$ax^2+bx+c=0$$ is equal to $$-\ \frac{\ b}{a}$$

3p = $$\frac{3}{2}$$

p = $$\frac{1}{2}$$

2p = 1

Product of roots of a quadratic equation $$ax^2+bx+c=0$$ is equal to $$\frac{c}{a}$$

2$$p^2$$ =$$\ \dfrac{\ 1}{2}$$

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

c = $$\ \dfrac{\ 1}{\sqrt{\ 2}}$$

Now, we have the equation  $$a^{2}x^{2} + 12a - 7 = 0$$ having a root equal to $$64\left(p^{6}+c^{12}\right)$$ 

$$64\left(p^{6}+c^{12}\right)$$ = $$64\left(\ \dfrac{1}{64}+\dfrac{1}{64}\right)$$

= 2

So x=2 would be a root of the equation $$a^{2}x^{2} + 12a - 7 = 0$$

Putting the value of x=2 in the equation, we get,

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

a=$$\dfrac{\left(-12\pm\ \sqrt{\ 144+4\cdot4\cdot7}\right)}{8}$$

= $$\dfrac{1}{2}$$ , -$$\dfrac{7}{2}$$

Option b is the correct answer.