Let $$A = \{1, 2, 3, 4, 5, 6, 7\}$$. Then the relation $$R = \{(x, y) \in A \times A : x + y = 7\}$$ is

$$A = \{1, 2, 3, 4, 5, 6, 7\}$$ and $$R = \{(x, y) \in A \times A : x + y = 7\}$$

The pairs in $$R$$ are: $$(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$$

Reflexive? For reflexivity, $$(a, a) \in R$$ for all $$a \in A$$. This requires $$a + a = 7$$, i.e., $$2a = 7$$, giving $$a = 3.5 \notin A$$. So $$(1,1), (2,2), \ldots$$ are not in $$R$$. Not reflexive.

Symmetric? If $$(x, y) \in R$$, then $$x + y = 7$$, which means $$y + x = 7$$, so $$(y, x) \in R$$. We can verify: $$(1,6) \in R$$ and $$(6,1) \in R$$, $$(2,5) \in R$$ and $$(5,2) \in R$$, etc. Symmetric.

Transitive? If $$(x, y) \in R$$ and $$(y, z) \in R$$, then $$x + y = 7$$ and $$y + z = 7$$, so $$x = 7 - y$$ and $$z = 7 - y$$, giving $$x = z$$. For transitivity we need $$(x, z) = (x, x) \in R$$, which requires $$2x = 7$$. Since $$x$$ is an integer, this is impossible. For example, $$(1,6) \in R$$ and $$(6,1) \in R$$, but $$(1,1) \notin R$$. Not transitive.

Therefore, $$R$$ is symmetric but neither reflexive nor transitive, which is Option B.