To solve the problem, we start by defining the set $A = \{-2, -1, 0, 1, 2, 3\}$ and the relation $R$ on set $A$, where an element $x$ is related to $y$ (written as $x \, R \, y$) if and only if $y = \max\{x, 1\}$.

This leads us to the following pairs in the relation $R$:

For $x = -2$, $y = \max\{-2, 1\} = 1$, so $(-2, 1)$ is in $R$.

For $x = -1$, $y = \max\{-1, 1\} = 1$, so $(-1, 1)$ is in $R$.

For $x = 0$, $y = \max\{0, 1\} = 1$, so $(0, 1)$ is in $R$.

For $x = 1$, $y = \max\{1, 1\} = 1$, so $(1, 1)$ is in $R$.

For $x = 2$, $y = \max\{2, 1\} = 2$, so $(2, 2)$ is in $R$.

For $x = 3$, $y = \max\{3, 1\} = 3$, so $(3, 3)$ is in $R$.

Thus, the relation $R$ consists of the pairs: $\{(-2, 1), (-1, 1), (0, 1), (1, 1), (2, 2), (3, 3)\}$, and there are $l = 6$ elements in $R$.

Making the Relation Reflexive

A relation is reflexive if every element in the set $A$ relates to itself. Therefore, the missing reflexive pairs are:

$(-2, -2)$

$(-1, -1)$

$(0, 0)$

Adding these three pairs will make the relation reflexive, so $m = 3$.

Making the Relation Symmetric

A relation is symmetric if whenever $(x, y)$ is in $R$, $(y, x)$ must also be in $R$. Therefore, the missing symmetric pairs are:

$(1, -2)$

$(1, -1)$

$(1, 0)$

Thus, we need to add these three pairs for symmetry, so $n = 3$.

Finally, we calculate the sum $l + m + n = 6 + 3 + 3 = 12$.