Answer
See below
Work Step by Step
We can write set $S$ as $S=\{A =\begin{bmatrix}
a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32}
\end{bmatrix} \in M_2(R): a_{11}+a_{21}+a_{31}=0,a_{12}+a_{22}+a_{32}=0\}$.
We can notice that $\begin{bmatrix}
0 & 0 \\ 0 & 0 \\ 0 & 0
\end{bmatrix} \in S$, hence $S$ is nonempty set.
A1. Let $A=\begin{bmatrix}
a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32}
\end{bmatrix} ,B=\begin{bmatrix}
b_{11} & b_{12} \\ b_{21} & b_{22} \\ b_{31} & b_{32}
\end{bmatrix} \in S$
Obtain: $A+B=\begin{bmatrix}
a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32}
\end{bmatrix} +\begin{bmatrix}
b_{11} & b_{12} \\ b_{21} & b_{22} \\ b_{31} & b_{32}
\end{bmatrix} =\begin{bmatrix}
a_{11}+b_{11} & a_{12}+b_{12} \\ a_{21}+b_{21} & a_{22}+b_{22} \\ a_{31}+b_{31} & a_{32}+b_{32}
\end{bmatrix} =0$
Since $A+B \in S$, set $S$ is closed under addition.
A2. Let $A=\begin{bmatrix}
a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32}
\end{bmatrix} \in S$ and $k \in R$
Obtain: $kA=k\begin{bmatrix}
a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32}
\end{bmatrix} =\begin{bmatrix}
ka_{11} & ka_{12} \\ ka_{21} & ka_{22} \\ka_{31} & ka_{32}
\end{bmatrix} =k.0=0$
Since $kA \in S$, set $S$ is closed under scalar multiplication (2)
From (1),(2) and since $S$ is nonempty subset, $S$ is a subspace of $S$
