Let the focal chord PQ of the parabola $$y^2 = 4x$$ with the positive x-axis, make an angle of $$60^\circ$$ where P lies in the first quadrant. If the circle, whose one diameter is PS, S being the focus of the parabola, touches the y-axis at the point $$(0, \alpha)$$, then $$5\alpha^2$$ is equal to:

The given parabola is $$y^{2}=4x$$.
For $$y^{2}=4ax$$ we have $$a=1$$, so its focus is $$S(1,0)$$.

Let the focal chord $$PQ$$ pass through the focus $$S$$ and make an angle $$60^{\circ}$$ with the positive $$x$$-axis.
Hence the slope of $$PQ$$ is $$\tan 60^{\circ}= \sqrt{3}$$.

Equation of the chord through $$S(1,0)$$ with this slope:
$$y-0=\sqrt{3}(x-1)\; \Longrightarrow\; y=\sqrt{3}(x-1) \; -(1)$$

The points $$P(x_1,y_1)$$ and $$Q(x_2,y_2)$$ lie on the parabola, so substitute $$y$$ from $$(1)$$ into $$y^{2}=4x$$:

$$\bigl[\sqrt{3}(x-1)\bigr]^{2}=4x$$
$$3(x-1)^{2}=4x$$
$$3x^{2}-6x+3-4x=0$$
$$3x^{2}-10x+3=0 \; -(2)$$

Solving $$(2)$$:
Discriminant $$D=(-10)^{2}-4\cdot3\cdot3=100-36=64$$.
$$x=\dfrac{10\pm8}{2\cdot3}=\dfrac{10\pm8}{6}$$ gives
$$x_1=3,\; x_2=\dfrac{1}{3}$$.

Since point $$P$$ lies in the first quadrant, we take $$x_1=3$$.
Using $$(1)$$, $$y_1=\sqrt{3}(3-1)=2\sqrt{3}$$.
Thus $$P(3,\,2\sqrt{3})$$.

Consider the circle having $$PS$$ as a diameter.
If two points $$A(x_1,y_1)$$ and $$B(x_2,y_2)$$ are the ends of a diameter, the centre is the midpoint $$M\bigl(\tfrac{x_1+x_2}{2},\tfrac{y_1+y_2}{2}\bigr)$$ and the radius is $$\tfrac{1}{2}AB$$.

For $$P(3,2\sqrt{3})$$ and $$S(1,0)$$:
Centre $$M\left(\dfrac{3+1}{2},\dfrac{2\sqrt{3}+0}{2}\right)=(2,\sqrt{3})$$.
Radius $$R=\dfrac{1}{2}\sqrt{(3-1)^{2}+(2\sqrt{3}-0)^{2}}=\dfrac{1}{2}\sqrt{4+12}=2$$.

The circle is to touch the $$y$$-axis.
For a circle with centre $$(h,k)$$ and radius $$R$$, tangency to the line $$x=0$$ (the $$y$$-axis) requires $$|h|=R$$ and the point of contact is $$(0,k)$$.

Here $$h=2$$ and $$R=2$$, so the condition is satisfied.
Hence the touching point is $$(0,\alpha)=(0,k)=(0,\sqrt{3})$$, giving $$\alpha=\sqrt{3}$$.

Finally,
$$5\alpha^{2}=5(\sqrt{3})^{2}=5\cdot3=15$$.

Therefore, the value of $$5\alpha^{2}$$ is $$15$$, which corresponds to Option A.