Here,

${R_1} = \left\{ {\left( {x,y} \right) \in N \times N:2x + y = 10} \right\}$

${R_2} = \left\{ {\left( {x,y} \right) \in N \times N:x + 2y = 10} \right\}$

For ${R_1};2x + y = 10$ and $x,y \in N$

So, possible values for $x$ and $y$ are:

$x = 1,y = 8$ i.e. $(1,8);$

$x = 2,y = 6\,$ i.e. $(2,6);$

$x = 3,y = 4$ i.e. $\,(3,4)$ and 

$x = 4,y = 2\,$ i.e. $(4,2)$.

${R_1} = \left\{ {(1,8),(2,6),(3,4),(4,2)} \right\}$

Therefore, Range of ${R_1}$ is $\left\{ {2,4,6,8} \right\}$

${R_1}$ is not symmetric

Also,${R_1}$ is not transitive because

$(3,4),(4,2) \in {R_1}$ but $\,(3,2) \notin {R_1}$

Thus, options $A,B$ and $D$ are incorrect.

For ${R_2};x + 2y = 10$ and $x,y \in N$

So, possible values for $x$ and $y$ are:

$x = 8,y = 1\,$i.e.$\,(8,1);$

$x = 6,y = 2$i.e.$(6,2);$

$x = 4,y = 3$i.e.$(4,3)$ and 

$x = 2,y = 4$i.e.$(2,4)$

${R_2} = \left\{ {(8,1),(6,2),(4,3),(2,4)} \right\}$

Therefore, Range of ${R_2}$ is $\left\{ {1,2,3,4} \right\}$

 ${R_2}$ is not symmetric and transitive.