Let $$P(\alpha, \beta)$$ be a point on the parabola $$y^2 = 4x$$. If $$P$$ also lies on the chord of the parabola $$x^2 = 8y$$ whose mid point is $$\left(1, \frac{5}{4}\right)$$, then $$(\alpha - 28)(\beta - 8)$$ is equal to ______.

Point $$P(\alpha, \beta)$$ lies on the parabola $$y^2 = 4x$$, so $$\beta^2 = 4\alpha$$.

$$P$$ also lies on the chord of the parabola $$x^2 = 8y$$ whose midpoint is $$\left(1, \frac{5}{4}\right)$$.

Find the equation of the chord: for the parabola $$x^2 = 8y$$, the equation of the chord with midpoint $$(h, k)$$ is given by $$T = S_1$$:

$$xh - 4(y + k) = h^2 - 8k$$

With $$(h, k) = \left(1, \frac{5}{4}\right)$$:

$$x(1) - 4\left(y + \frac{5}{4}\right) = 1 - 8 \cdot \frac{5}{4}$$

$$x - 4y - 5 = 1 - 10 = -9$$

$$x - 4y + 4 = 0 \implies x = 4y - 4$$

Use both conditions to find $$\alpha$$ and $$\beta$$: from the chord equation: $$\alpha = 4\beta - 4$$.

From the parabola: $$\beta^2 = 4\alpha = 4(4\beta - 4) = 16\beta - 16$$.

$$ \beta^2 - 16\beta + 16 = 0 $$

$$ \beta = \frac{16 \pm \sqrt{256 - 64}}{2} = \frac{16 \pm \sqrt{192}}{2} = \frac{16 \pm 8\sqrt{3}}{2} = 8 \pm 4\sqrt{3} $$

Compute $$(\alpha - 28)(\beta - 8)$$: if $$\beta = 8 + 4\sqrt{3}$$: $$\alpha = 4(8 + 4\sqrt{3}) - 4 = 28 + 16\sqrt{3}$$.

$$(\alpha - 28)(\beta - 8) = (16\sqrt{3})(4\sqrt{3}) = 64 \times 3 = 192$$

If $$\beta = 8 - 4\sqrt{3}$$: $$\alpha = 4(8 - 4\sqrt{3}) - 4 = 28 - 16\sqrt{3}$$.

$$(\alpha - 28)(\beta - 8) = (-16\sqrt{3})(-4\sqrt{3}) = 64 \times 3 = 192$$

In both cases, the answer is $$\boxed{192}$$.