Let $$A={(x,y) \in R\times R : |x+y|\geq 3}$$ and $$B={(x,y) \in R\times R : |x|+|y|\leq 3}$$. If $$C = \{(x,y) \in A \cap B : x = 0 \text{ or } y = 0\}$$,then $$\sum_{(x,y) \in C} |x+y|$$ is :

We need to find the points in $$A \cap B$$ where x=0 or y=0, and then compute $$\sum |x+y|$$ for those points.

The sets are defined by $$A = \{(x,y) \in \mathbb{R} \times \mathbb{R} : |x+y| \geq 3\}$$, $$B = \{(x,y) \in \mathbb{R} \times \mathbb{R} : |x|+|y| \leq 3\}$$, and we consider $$C = \{(x,y) \in A \cap B : x = 0 \text{ or } y = 0\}$$.

If x=0, then from A we have $$|y| \geq 3$$ and from B we have $$|y| \leq 3$$, which together imply $$|y| = 3$$, so y=3 or y=-3, giving the points $$(0,3)$$ and $$(0,-3)$$.

If y=0, then from A we have $$|x| \geq 3$$ and from B we have $$|x| \leq 3$$, which together imply $$|x| = 3$$, so x=3 or x=-3, giving the points $$(3,0)$$ and $$(-3,0)$$.

Thus $$C = \{(0,3), (0,-3), (3,0), (-3,0)\}$$.

It follows that $$\sum_{(x,y) \in C} |x+y| = |0+3| + |0-3| + |3+0| + |-3+0| = 3 + 3 + 3 + 3 = 12.$$

The correct answer is Option 4: 12.