If 2 and 6 are the roots of the equation $$ax^2 + bx + 1 = 0$$, then the quadratic equation whose roots are $$\frac{1}{2a+b}$$ and $$\frac{1}{6a+b}$$ is:

Let the given quadratic be $$ax^2 + bx + 1 = 0$$.

Because 2 and 6 are its roots, each satisfies the equation.

Substituting $$x = 2$$ gives $$4a + 2b + 1 = 0$$ $$-(1)$$

Substituting $$x = 6$$ gives $$36a + 6b + 1 = 0$$ $$-(2)$$

Subtract $$(1)$$ from $$(2)$$:

$$32a + 4b = 0 \;\;\Longrightarrow\;\; 8a + b = 0$$ $$-(3)$$

From $$(3)$$, $$b = -8a$$. Insert this value of $$b$$ into $$(1)$$:

$$4a + 2(-8a) + 1 = 0 \;\;\Longrightarrow\;\; 4a - 16a + 1 = 0$$

$$-12a + 1 = 0 \;\;\Longrightarrow\;\; a = \frac{1}{12}$$

Using $$b = -8a$$, we get $$b = -\frac{8}{12} = -\frac{2}{3}$$.

Next, find $$2a + b$$ and $$6a + b$$:

$$2a + b = 2\left(\frac{1}{12}\right) - \frac{2}{3} = \frac{1}{6} - \frac{2}{3} = -\frac{1}{2}$$

$$6a + b = 6\left(\frac{1}{12}\right) - \frac{2}{3} = \frac{1}{2} - \frac{2}{3} = -\frac{1}{6}$$

The required new roots are $$\frac{1}{2a+b}$$ and $$\frac{1}{6a+b}$$:

$$\frac{1}{2a+b} = \frac{1}{-\frac{1}{2}} = -2$$

$$\frac{1}{6a+b} = \frac{1}{-\frac{1}{6}} = -6$$

Hence the new quadratic, with roots $$-2$$ and $$-6$$, is obtained by

$$(x + 2)(x + 6) = 0$$

Multiplying out gives $$x^2 + 8x + 12 = 0$$.

Therefore, the required quadratic equation is $$x^2 + 8x + 12 = 0$$, which corresponds to Option B.