The sum of the square of the modulus of the elements in the set $$\{z = a + ib : a, b \in \mathbb{Z}, z \in \mathbb{C}, |z - 1| \leq 1, |z - 5| \leq |z - 5i|\}$$ is ________

We need $$z = a + ib$$ with $$a, b \in \mathbb{Z}$$ satisfying: $$|z-1| \leq 1$$ and $$|z-5| \leq |z-5i|$$.

Condition 1: $$|z-1| \leq 1$$ means $$(a-1)^2 + b^2 \leq 1$$.

Since $$a, b$$ are integers: possible points are $$(0,0), (1,0), (2,0), (1,1), (1,-1)$$.

Check: $$(0,0)$$: $$(0-1)^2+0 = 1 \leq 1$$ ✓. $$(1,0)$$: $$0 \leq 1$$ ✓. $$(2,0)$$: $$1 \leq 1$$ ✓. $$(1,1)$$: $$0+1=1 \leq 1$$ ✓. $$(1,-1)$$: $$0+1=1 \leq 1$$ ✓.

Condition 2: $$|z-5| \leq |z-5i|$$, i.e., $$(a-5)^2+b^2 \leq a^2+(b-5)^2$$.

$$a^2-10a+25+b^2 \leq a^2+b^2-10b+25$$

$$-10a \leq -10b \implies b \leq a$$.

From the 5 points, those with $$b \leq a$$:

$$(0,0)$$: $$0 \leq 0$$ ✓. $$(1,0)$$: $$0 \leq 1$$ ✓. $$(2,0)$$: $$0 \leq 2$$ ✓. $$(1,1)$$: $$1 \leq 1$$ ✓. $$(1,-1)$$: $$-1 \leq 1$$ ✓.

All 5 points satisfy both conditions.

Sum of $$|z|^2$$: $$0 + 1 + 4 + 2 + 2 = 9$$.

The answer is $$\boxed{9}$$.