JEE Main 2025 Jan 23 Shift 2, Mathematics Q2: Set

$B$ is the closed diamond with vertices $(\pm 3,0)$ and $(0,\pm 3)$. $A$ is the complement of the open strip $|x+y|<3$.

Inside $B$, the lines $x+y=\pm 3$ pass through the vertices and form two sides of the diamond. So $A\cap B$ consists only of the two triangular sides of the diamond where $|x+y|\ge 3$ holds with equality on the boundary.

Restricting further to points with $x=0$ or $y=0$ (the axes) keeps only the four diamond vertices: $C=\{(3,0),(-3,0),(0,3),(0,-3)\}$.

At each of these four vertices $|x+y|=3$, so the sum is $4\times 3=12$.