Let $$PQ$$ be a focal chord of the parabola $$y^2 = 4x$$ such that it subtends an angle of $$\frac{\pi}{2}$$ at the point $$(3, 0)$$. Let the line segment $$PQ$$ be also a focal chord of the ellipse $$E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$, $$a^2 > b^2$$. If $$e$$ is the eccentricity of the ellipse $$E$$, then the value of $$\frac{1}{e^2}$$ is equal to

Step 1: Set up the focal chord of parabola $$y^2 = 4x$$.

The focus of $$y^2 = 4x$$ is at $$S(1, 0)$$. A parametric point on the parabola is $$(t^2, 2t)$$. For a focal chord, if one end is $$P = (t^2, 2t)$$, the other end is $$Q = (1/t^2, -2/t)$$.

Step 2: Apply the condition that PQ subtends $$\frac{\pi}{2}$$ at $$(3, 0)$$.

Slope from $$(3, 0)$$ to $$P = (t^2, 2t)$$: $$m_1 = \frac{2t}{t^2 - 3}$$

Slope from $$(3, 0)$$ to $$Q = (1/t^2, -2/t)$$: $$m_2 = \frac{-2/t}{1/t^2 - 3} = \frac{-2/t}{\frac{1 - 3t^2}{t^2}} = \frac{-2t}{1 - 3t^2} = \frac{2t}{3t^2 - 1}$$

For perpendicularity, $$m_1 \cdot m_2 = -1$$:

$$\frac{2t}{t^2 - 3} \cdot \frac{2t}{3t^2 - 1} = -1$$

$$\frac{4t^2}{(t^2 - 3)(3t^2 - 1)} = -1$$

$$4t^2 = -(3t^4 - 10t^2 + 3)$$

$$3t^4 - 10t^2 + 3 + 4t^2 = 0$$

$$3t^4 - 6t^2 + 3 = 0 \implies t^4 - 2t^2 + 1 = 0 \implies (t^2 - 1)^2 = 0$$

So $$t^2 = 1$$, giving $$t = 1$$ (taking positive value).

Step 3: Identify the focal chord PQ.

$$P = (1, 2)$$ and $$Q = (1, -2)$$.

This is the latus rectum of the parabola with length $$PQ = 4$$.

Verification: Slope from $$(3,0)$$ to $$P(1,2)$$ is $$\frac{2}{-2} = -1$$; slope to $$Q(1,-2)$$ is $$\frac{-2}{-2} = 1$$. Product $$= -1$$. $$\checkmark$$

Step 4: Relate to the ellipse.

PQ is also a focal chord of the ellipse $$E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ with $$a^2 > b^2$$.

Since PQ passes through $$(1, 0)$$, the focus of the ellipse is at $$(1, 0)$$, so $$ae = 1$$.

The chord PQ is vertical ($$x = 1$$) and passes through the focus, so it is the latus rectum of the ellipse.

The length of the latus rectum of the ellipse is $$\frac{2b^2}{a}$$.

Setting this equal to 4:

$$\frac{2b^2}{a} = 4 \implies b^2 = 2a$$

Since $$b^2 = a^2 - a^2e^2 = a^2 - 1$$ (using $$ae = 1$$):

$$a^2 - 1 = 2a \implies a^2 - 2a - 1 = 0$$

$$a = \frac{2 + \sqrt{4 + 4}}{2} = 1 + \sqrt{2}$$

Step 5: Compute $$\frac{1}{e^2}$$.

Since $$ae = 1$$, we have $$\frac{1}{e} = a = 1 + \sqrt{2}$$.

$$\frac{1}{e^2} = a^2 = (1 + \sqrt{2})^2 = 1 + 2\sqrt{2} + 2 = 3 + 2\sqrt{2}$$

The answer is Option B: $$3 + 2\sqrt{2}$$.