The area bounded by the lines $$y = ||x - 1| - 2|$$ and $$y = 2$$ is ______.

We need to find the area bounded by $$y = ||x-1| - 2|$$ and $$y = 2$$.

Let us analyze $$y = ||x-1| - 2|$$. Setting $$u = |x-1|$$, we have $$y = |u - 2|$$. The graph of $$y = |x-1|$$ is a V-shape with vertex at $$(1, 0)$$. Then $$|x-1| - 2$$ shifts it down by 2, creating a W-shape after taking the absolute value.

The function $$y = ||x-1| - 2|$$ equals 0 when $$|x-1| = 2$$, i.e., at $$x = 3$$ and $$x = -1$$. It equals 2 when $$|x-1| - 2 = \pm 2$$, giving $$|x-1| = 0$$ (so $$x = 1$$) or $$|x-1| = 4$$ (so $$x = 5$$ or $$x = -3$$).

The curve $$y = ||x-1| - 2|$$ meets $$y = 2$$ at $$x = -3, 1, 5$$. Between these intersection points, the curve lies below $$y = 2$$.

By symmetry about $$x = 1$$, the area is $$2 \int_{-3}^{1} (2 - ||x-1| - 2|) \, dx$$. Using the substitution $$t = 1 - x$$ (or equivalently working on $$[1, 5]$$):

$$\text{Area} = 2\int_1^5 (2 - ||x-1| - 2|) \, dx = 2\int_0^4 (2 - |t - 2|) \, dt$$

where $$t = x - 1$$. For $$t \in [0, 2]$$: $$|t-2| = 2-t$$, so the integrand is $$2 - (2-t) = t$$. For $$t \in [2, 4]$$: $$|t-2| = t-2$$, so the integrand is $$2 - (t-2) = 4-t$$.

$$2\left[\int_0^2 t \, dt + \int_2^4 (4-t) \, dt\right] = 2\left[\frac{t^2}{2}\Big|_0^2 + \left(4t - \frac{t^2}{2}\right)\Big|_2^4\right]$$

$$= 2\left[2 + (16 - 8) - (8 - 2)\right] = 2\left[2 + 8 - 6\right] = 2 \times 4 = 8$$.

The area bounded by the curves is $$8$$.