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.

If you subtract the second equation from the first you have that 
$x-x+5y-3y+z-2z=0$
So
$z=2y$ 
Now if you change $z$ with $2y$ in the third equation you have that 
$3x+5y+16y=0$
$x=-7y$
Now if you change $x$ with $-7y$ in the first equation you have that 
$-7y+3y+4y=0$ for all $y\in \mathbb{R}$ 
So the set of solution of your system will be 
$W=\{\lambda (-7, 1,2 ): \lambda \in \mathbb{R}\}$ 
So a base of your subspace will be $\{(-7,1,2)\}$
