It's better to write the subset as:
$$W=\left\{\begin{bmatrix}
3s - 2t \\
s + 2t \\
2s + 3t \\
\end{bmatrix}:s,t \in \mathbb{R}\right\}.$$
I know that the conditions for a subset being a subspace is that it most include the zero vector, and be closed under addition and multiplication. However, I do not know how to do this.
To show the zero vector belongs to the set, we find values of $s \in \mathbb{R}$ and $t \in \mathbb{R}$ such that
$$\begin{bmatrix}
3s - 2t \\
s + 2t \\
2s + 3t \\
\end{bmatrix}=\begin{bmatrix}
0 \\
0 \\
0 \\
\end{bmatrix}.$$
To show that it's closed under addition, we take two arbitrary elements from $W$, add them together, and show that the result is in $W$. I.e. show
$$
\begin{bmatrix}
3s - 2t \\
s + 2t \\
2s + 3t \\
\end{bmatrix}
+
\begin{bmatrix}
3s' - 2t' \\
s' + 2t' \\
2s' + 3t' \\
\end{bmatrix}
=
\begin{bmatrix}
3(s+s') - 2(t+t') \\
(s+s') + 2(t+t') \\
2(s+s') + 3(t+t') \\
\end{bmatrix}
$$
belongs to $W$.
To show that it's closed under scalar multiplication, we take an arbitrary element from $W$, and multiply it by an arbitrary scalar $\alpha \in \mathbb{R}$, and show that the result is in $W$. I.e., we show
$$
\alpha
\begin{bmatrix}
3s - 2t \\
s + 2t \\
2s + 3t \\
\end{bmatrix}
=
\begin{bmatrix}
3\alpha s - 2\alpha t \\
\alpha s + 2\alpha t \\
2\alpha s + 3\alpha t \\
\end{bmatrix}
$$
is in $W$.
To identify a basis (a spanning, linearly independent subset), the co-efficients of $s$ and the coefficients of $t$ should give hint. We can rewrite $W$ as:
$$W=\left\{s\begin{bmatrix}
3 \\
1 \\
2 \\
\end{bmatrix}+t\begin{bmatrix}
- 2 \\
2 \\
3 \\
\end{bmatrix}:s,t \in \mathbb{R}\right\}.$$ Note: we need to check that the basis vectors are indeed linearly independent.
The dimension of $W$ is the size of any of it's bases, by the Dimension Theorem, so once you've found a basis, the answer to this part is "what's it's size?".
