If the $$x$$-intercept of a focal chord of the parabola $$y^2 = 8x + 4y + 4$$ is $$3$$, then the length of this chord is equal to ______.

We need to find the length of a focal chord of the parabola $$y^2 = 8x + 4y + 4$$ whose x-intercept is 3.

Rewriting the given equation as $$y^2 - 4y = 8x + 4$$ and completing the square gives $$(y - 2)^2 = 8(x + 1)$$, which is of the standard form $$Y^2 = 8X$$ with the substitutions $$Y = y - 2$$ and $$X = x + 1$$.

Since $$4a = 8$$, it follows that $$a = 2$$, the vertex of the parabola is at $$(-1,2)$$, and the focus is at $$(1,2)$$.

A focal chord must pass through the focus at $$(1,2)$$ and also through the point of x-intercept $$(3,0)$$. The slope of this line is $$m = \frac{2 - 0}{1 - 3} = -1$$, so its equation is $$y - 2 = -1 \cdot (x - 1)$$ or equivalently $$y = 3 - x$$.

Substituting $$y = 3 - x$$ into the standard form $$(y - 2)^2 = 8(x + 1)$$ gives $$(3 - x - 2)^2 = 8(x + 1)$$, which simplifies to $$(1 - x)^2 = 8x + 8$$.

Expanding and rearranging yields $$x^2 - 2x + 1 = 8x + 8$$ and hence $$x^2 - 10x - 7 = 0$$, whose solutions are $$x = \frac{10 \pm \sqrt{100 + 28}}{2} = 5 \pm 4\sqrt{2}$$.

The corresponding y-values follow from $$y = 3 - x$$, giving $$y_1 = 3 - (5 + 4\sqrt{2}) = -2 - 4\sqrt{2}$$ and $$y_2 = 3 - (5 - 4\sqrt{2}) = -2 + 4\sqrt{2}$$.

The horizontal and vertical differences between these two points are $$\Delta x = (5 + 4\sqrt{2}) - (5 - 4\sqrt{2}) = 8\sqrt{2}$$ and $$\Delta y = (-2 - 4\sqrt{2}) - (-2 + 4\sqrt{2}) = -8\sqrt{2}$$.

Therefore, the length of the chord is $$\sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(8\sqrt{2})^2 + (-8\sqrt{2})^2} = \sqrt{128 + 128} = \sqrt{256} = 16$$.