$$\alpha $$ and $$\beta $$ are the roots of x² $$-$$ 3x + p = 0

$$ \therefore $$ $$\alpha $$ + $$\beta $$ = 3 and $$\alpha \beta $$ = p

$$\gamma $$ and $$\delta $$ are the roots of x² $$-$$ 6x + q = 0

$$ \therefore $$ $$\gamma $$ + $$\delta $$ = 6 and $$\gamma \delta $$ = q

Given that, $$\alpha $$, $$\beta $$, $$\gamma $$, $$\delta $$ are in G.P.

$$ \therefore $$ Let $$\alpha $$ = a, $$\beta $$ = ar, $$\gamma $$ = ar² and $$\delta $$ = ar³

As $$\alpha $$ + $$\beta $$ = 3

$$ \Rightarrow $$ a + ar = 3

$$ \Rightarrow $$ a(1 + r) = 3 ...(1)

Also $$\gamma $$ + $$\delta $$ = 6

$$ \Rightarrow $$ ar² + ar³ = 6

$$ \Rightarrow $$ ar² (1 + r) = 6 ... (2)

Dividing (2) by (1),

r² = 2

$$ \Rightarrow r = \sqrt 2 $$

$$ \therefore $$ $$a = {3 \over {1 + \sqrt 2 }}$$

So, $$\alpha = {3 \over {1 + \sqrt 2 }}$$, $$\beta = {{3\sqrt 2 } \over {1 + \sqrt 2 }}$$, $$\gamma = {{3 \times 2} \over {1 + \sqrt 2 }}$$, $$\delta = {{3(2\sqrt {2)} } \over {1 + \sqrt 2 }}$$

$$ \therefore $$ $$p = \alpha \beta = {{9\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}}$$

and $$q = \gamma \delta = {{36\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}}$$

Now, $${{2q + p} \over {2q - p}}$$

$$ = {{{{72\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}} + {{9\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}}} \over {{{72\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}} - {{9\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}}}}$$

$$ = {{72\sqrt 2 + 9\sqrt 2 } \over {72\sqrt 2 - 9\sqrt 2 }}$$

$$ = {{81} \over {63}}$$

$$ = {9 \over 7}$$