Answer
See below
Work Step by Step
Given: $V=M_{3\times 2}(R)\\
S=\{\begin{bmatrix}
a & b \\ c & d \\ e &f
\end{bmatrix}\in M_{3\times 2}(R):a+b=c+f,a-c=e-f-d\}$
We can see $O=\begin{bmatrix}
0 & 0 \\ 0 & 0 \\ 0 & 0
\end{bmatrix} \in W$, hence $W$ is nonempty (1)
Let $A=\begin{bmatrix}
a & b \\ c & d \\ e &f
\end{bmatrix}\\B=\begin{bmatrix}
a' & b' \\ c' & d' \\ e' &f'
\end{bmatrix}$
then $A+B=\begin{bmatrix}
a & b \\ c & d \\ e &f
\end{bmatrix}+\begin{bmatrix}
a' & b' \\ c' & d' \\ e' &f'
\end{bmatrix}=\begin{bmatrix}
a+a' & b+b' \\ c+c' & d+d' \\ e+e' &f+f'
\end{bmatrix}\\
\rightarrow A+B\in W$
Hence, $A+B$ is closed under addition multiplication (2)
Let $k$ be a scalar
Obtain $k\begin{bmatrix}
a & b \\ c & d \\ e &f
\end{bmatrix}\\
=k(a+b)=k(c+f)$
and $k(a-c)=k(e-f-d)$
$\rightarrow ka+kb=kc+kf$ or $ka-kc=ke-kf-kd$
$\rightarrow kA=\begin{bmatrix}
ka & kb \\ kc & kd \\ ke &kf
\end{bmatrix} \in W$
Hence, $kA$ is closed under scalar multiplication (3)
From (1)(2)(3), $W$ is a subspace of $M_{2\times 3}$
