Let A={-3,-2,-1,0,1,2,3} and R be a relation on A defined by xRy iff 2x-y $\in\{0,1\}$. Let $l$ be the number of elements in R. Let m and n be the minimum number of elements required to be added in R to make it reflexive and symmetric relations, respectively. Then l+m+n is equal to

Given $A=\{-3,-2,-1,0,1,2,3\}$ and $xRy \iff 2x-y\in\{0,1\}$

So $y=2x$ or $y=2x-1$

All valid pairs in $A$:
$(-1,-2),\ (-1,-3),\ (0,0),\ (0,-1),\ (1,2),\ (1,1),\ (2,3)$

So $l=7$

Reflexive pairs needed: $(x,x)$ for all 7 elements
Present: $(0,0),(1,1)$ ⇒ missing $5$ ⇒ $m=5$

For symmetry, add reverse of non-symmetric pairs:
Missing: $(-2,-1),\ (-3,-1),\ (-1,0),\ (2,1),\ (3,2)$ ⇒ $n=5$

$l+m+n=7+5+5=17$

$\boxed{17}$