If $$\alpha$$ and $$\beta$$ are roots of the equation $$x^2 + px + \frac{3p}{4} = 0$$, such that $$|\alpha - \beta| = \sqrt{10}$$, then $$p$$ belongs to the set :

We are given the quadratic equation

$$x^{2}+px+\frac{3p}{4}=0$$

and we are told that its roots are $$\alpha$$ and $$\beta$$ with

$$|\alpha-\beta|=\sqrt{10}.$$

Because the quadratic is monic (the coefficient of $$x^{2}$$ is $$1$$), we can immediately quote the quadratic formula. For a general quadratic

$$x^{2}+bx+c=0,$$

the discriminant is

$$\Delta=b^{2}-4c.$$

The two roots are

$$\alpha=\frac{-b+\sqrt{\Delta}}{2},\qquad \beta=\frac{-b-\sqrt{\Delta}}{2}.$$

The numerical distance between the roots is therefore

$$|\alpha-\beta|=\left|\frac{-b+\sqrt{\Delta}}{2}-\frac{-b-\sqrt{\Delta}}{2}\right| =\left|\frac{2\sqrt{\Delta}}{2}\right| =\sqrt{\Delta}.$$

So for a monic quadratic, the formula simplifies to

$$|\alpha-\beta|=\sqrt{\Delta}.$$

In our particular equation the role of $$b$$ is played by $$p$$ and the role of $$c$$ by $$\dfrac{3p}{4}$$. Hence the discriminant is

$$\Delta=p^{2}-4\left(\frac{3p}{4}\right)=p^{2}-3p.$$

Using the given magnitude of the root difference, we set

$$|\alpha-\beta|=\sqrt{10}\;\Longrightarrow\;\sqrt{\Delta}=\sqrt{10}.$$

Squaring both sides gives

$$\Delta=10.$$

Substituting the expression for $$\Delta$$, we get

$$p^{2}-3p=10.$$

Re-arranging all terms to one side:

$$p^{2}-3p-10=0.$$

This is a quadratic in $$p$$. We solve it by factoring or by the quadratic formula. Using the quadratic formula

$$p=\frac{3\pm\sqrt{(-3)^{2}-4(1)(-10)}}{2\cdot1} =\frac{3\pm\sqrt{9+40}}{2} =\frac{3\pm\sqrt{49}}{2} =\frac{3\pm7}{2}.$$

This yields two values:

$$p=\frac{3+7}{2}=5,\qquad p=\frac{3-7}{2}=-2.$$

Thus the parameter $$p$$ can take the values $$-2$$ or $$5$$, i.e.

$$p\in\{-2,\,5\}.$$

Comparing with the given options, we see that this set corresponds exactly to Option C.

Hence, the correct answer is Option C.