Let $$\lambda$$ be an integer. If the shortest distance between the lines $$x - \lambda = 2y - 1 = -2z$$ and $$x = y + 2\lambda = z - \lambda$$ is $$\frac{\sqrt{7}}{2\sqrt{2}}$$, then the value of $$|\lambda|$$ is ______.

We rewrite the two lines in standard symmetric form. The first line $$x - \lambda = 2y - 1 = -2z$$ can be written as $$\frac{x - \lambda}{1} = \frac{y - \frac{1}{2}}{\frac{1}{2}} = \frac{z - 0}{-\frac{1}{2}}$$, so it passes through $$\left(\lambda, \frac{1}{2}, 0\right)$$ with direction vector $$(2, 1, -1)$$ (after scaling).

The second line $$x = y + 2\lambda = z - \lambda$$ can be written as $$\frac{x}{1} = \frac{y + 2\lambda}{1} = \frac{z - \lambda}{1}$$, passing through $$(0, -2\lambda, \lambda)$$ with direction vector $$(1, 1, 1)$$.

The cross product of the direction vectors is $$(2, 1, -1) \times (1, 1, 1) = (1 \cdot 1 - (-1) \cdot 1,\; (-1) \cdot 1 - 2 \cdot 1,\; 2 \cdot 1 - 1 \cdot 1) = (2, -3, 1)$$, with magnitude $$\sqrt{4 + 9 + 1} = \sqrt{14}$$.

The vector connecting a point on each line is $$(0 - \lambda,\; -2\lambda - \frac{1}{2},\; \lambda - 0) = \left(-\lambda,\; -2\lambda - \frac{1}{2},\; \lambda\right)$$. Its dot product with $$(2, -3, 1)$$ is $$-2\lambda + 3\left(2\lambda + \frac{1}{2}\right) + \lambda = -2\lambda + 6\lambda + \frac{3}{2} + \lambda = 5\lambda + \frac{3}{2}$$.

The shortest distance is $$\frac{\left|5\lambda + \frac{3}{2}\right|}{\sqrt{14}} = \frac{\sqrt{7}}{2\sqrt{2}}$$. Solving: $$\left|5\lambda + \frac{3}{2}\right| = \frac{\sqrt{7} \cdot \sqrt{14}}{2\sqrt{2}} = \frac{\sqrt{98}}{2\sqrt{2}} = \frac{7\sqrt{2}}{2\sqrt{2}} = \frac{7}{2}$$.

So $$5\lambda + \frac{3}{2} = \pm\frac{7}{2}$$. This gives $$5\lambda = 2$$ (so $$\lambda = \frac{2}{5}$$) or $$5\lambda = -5$$ (so $$\lambda = -1$$). Since $$\lambda$$ must be an integer, $$\lambda = -1$$ and $$|\lambda| = 1$$.

The correct answer is $$1$$.