Find a basis and the dimension of the solution space $\textsf{W}$ $$\left\{\begin{align}
     x + 3y + 2z = 0 \\
     x + 5y + z  = 0 \\
     3x + 5y + 8z = 0 \\
\end{align}\right.$$
So if we represent this as an augmented matrix
$$\begin{pmatrix}
     1 & 3 & 2 & 0 \\
     1 & 5 & 1 & 0 \\
     3 & 5 & 8 & 0 \\
\end{pmatrix}$$
In row reduced form would be
$$\begin{pmatrix}
     1 & 0 & \tfrac{7}{2} & 0 \\
     0 & 1 & -\tfrac{1}{2} & 0 \\
     0 & 0 & 0 & 0 \\
\end{pmatrix}$$
Therefore, a basis of the set we can say would be vector
$$\begin{pmatrix} 1 \\ 1 \\ 3 \end{pmatrix}$$
and the vector $$\begin{pmatrix} 3 \\ 5 \\ 5 \end{pmatrix}$$
right? I'm new to these problems and want to make sure I've got the right idea in approaching the solution.

It is not correct. How did you obtain these basis vectors?
By row reduction you reduced your system to
$$\begin{cases} x_1 + \frac72x_3 = 0 \\ x_2 - \frac12x_3 = 0\end{cases}$$
so $$\begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix} = t\begin{bmatrix} -\frac72 \\ \frac12 \\ 1\end{bmatrix}, \quad\text{ for some } t \in \mathbb{R}$$
so e.g. $\left\{\begin{bmatrix} -7 \\ 1 \\ 2\end{bmatrix}\right\}$ is a basis for the solution space.
