$\mathrm{R} =\{( x , y ): y = x +5$ and $ x <4\}=\{(1,6),(2,7),(3,8)\}$

It is clear that $(1,1)\notin \mathrm{R}$

$\therefore $    $\mathrm{R}$ is not reflexive.

$(1,6) \in \mathrm{R}$ But, $(1,6)\notin \mathrm{R}$

$\therefore $   $\mathrm{R}$ is not symmetric.

Now, since there is no pair in $\mathrm{R}$ such that $( \mathrm{x} , \,\mathrm{y} )$ and $( \mathrm{y} ,\, \mathrm{z} ) \in \mathrm{R} ,$ then $( \mathrm{x} ,\, \mathrm{z} )$ cannot belong to $\mathrm{R}$.

$\therefore \mathrm{R}$ is not transitive.

Hence, $\mathrm{R}$ is neither reflexive, nor symmetric, nor transitive.