Let $$\alpha$$ and $$\beta$$ be the roots of $$x^2 - 3x + p = 0$$ and $$\gamma$$ and $$\delta$$ be the roots of $$x^2 - 6x + q = 0$$. If $$\alpha, \beta, \gamma, \delta$$ form a geometric progression. Then ratio $$(2q + p) : (2q - p)$$ is

For any quadratic of the form $$x^{2}-sx+t=0$$ we first recall the standard Vieta relations: the sum of its roots equals $$s$$ and the product equals $$t$$. Applying this fact to the two quadratics that are given, we have

For $$x^{2}-3x+p=0$$ :   $$\alpha+\beta=3,\qquad \alpha\beta=p.$$

For $$x^{2}-6x+q=0$$ :   $$\gamma+\delta=6,\qquad \gamma\delta=q.$$

It is stated that $$\alpha,\;\beta,\;\gamma,\;\delta$$ form a geometric progression. So, introducing the common ratio $$r,$$ we can write the four consecutive terms as

$$\alpha,\qquad \beta=\alpha r,\qquad \gamma=\alpha r^{2},\qquad \delta=\alpha r^{3}.$$

Now we put these expressions into the two sum-of-roots equations obtained above.

First, from $$\alpha+\beta=3$$ we have

$$\alpha+\alpha r=3 \;\;\Longrightarrow\;\; \alpha(1+r)=3 \;\;\Longrightarrow\;\; \alpha=\dfrac{3}{1+r}.$$

Second, from $$\gamma+\delta=6$$ we substitute $$\gamma=\alpha r^{2}$$ and $$\delta=\alpha r^{3}$$ to get

$$\alpha r^{2}+\alpha r^{3}=6 \;\;\Longrightarrow\;\; \alpha r^{2}(1+r)=6.$$

Replacing $$\alpha$$ by $$\dfrac{3}{1+r}$$ from the first relation, we obtain

$$\dfrac{3}{1+r}\,r^{2}(1+r)=6 \;\;\Longrightarrow\;\; 3r^{2}=6 \;\;\Longrightarrow\;\; r^{2}=2.$$

Thus $$r=\sqrt{2}$$ or $$r=-\sqrt{2}.$$ Either choice will eventually give the same ratio, because only even powers of $$r$$ will survive in the final expression. For definiteness we proceed with $$r=\sqrt{2}.$$

Next we find $$p=\alpha\beta$$. Using $$\beta=\alpha r,$$

$$p=\alpha\beta=\alpha(\alpha r)=\alpha^{2}r.$$

Since $$\alpha=\dfrac{3}{1+r},$$ this becomes

$$p=\left(\dfrac{3}{1+r}\right)^{2}r=\dfrac{9r}{(1+r)^{2}}.$$

To compute $$q=\gamma\delta,$$ we notice that

$$q=\gamma\delta=(\alpha r^{2})(\alpha r^{3})=\alpha^{2}r^{5}.$$

But $$\alpha^{2}r$$ has already appeared above as $$p,$$ so we write

$$q=p\,r^{4}.$$

Because $$r^{2}=2,$$ we have $$r^{4}=(r^{2})^{2}=2^{2}=4,$$ and therefore

$$q=4p.$$

Now we form the required ratio $$(2q+p):(2q-p).$$ First compute each linear combination in terms of $$p$$ only:

$$2q+p=2(4p)+p=8p+p=9p,$$

$$2q-p=2(4p)-p=8p-p=7p.$$

Hence

$$(2q+p):(2q-p)=9p:7p=9:7.$$

Therefore the desired ratio is $$9:7,$$ which corresponds to Option B.

Hence, the correct answer is Option B.