Check that $B$ is a basis $B= [a,b,c]$ where $a=[-3,2,1]^t$, $b=[-3,2,-1]^t$ and $c=[1,6,-1]^t$
How do I make this into a matrix to find if its a basis of $\mathbb{R}^3$?
I think the matrix will look like this but I'm not sure
$$\begin{bmatrix}
-3&-3&1\\
 2&2&6\\
 1&-1&1\\
\end{bmatrix}$$

You could also set
\begin{eqnarray*}
a&=&-3e_1+2e_2+e_3,\\
b&=&-3e_1+2e_2-e_3,\\
c&=&e_1+6e_2+e_3,
\end{eqnarray*}
to indicate the basis change from the old basis $\{e_1,e_2,e_3\}$ to the new one $\{a,b,c\}$.
Visually, the above table allows to find the new components of a vector, say, 
$$X=8e_1+199e_2-10e_3,$$
first you solve for the $e_i$
\begin{eqnarray*}
e_1&=&\frac{-1}{5}a+\frac{-1}{10}b+\frac{1}{10}c,\\
e_2&=&\frac{-1}{20}a+\frac{1}{10}b+\frac{3}{20}c,\\
e_3&=&\frac{1}{2}a+\frac{-1}{2}b.
\end{eqnarray*}  
With them, subbed in $x$, we're going to give its new components.
This correspond to a matricially multiplication    
$$
\left(
\begin{array}{ccc}
 -\frac{1}{5} & -\frac{1}{20} & \frac{1}{2} \\
 -\frac{1}{10} & \frac{1}{10} & -\frac{1}{2} \\
 \frac{1}{10} & \frac{3}{20} & 0 \\
\end{array}
\right)\cdot\left(
\begin{array}{c}
 8 \\
 199 \\
 -10 \\
\end{array}
\right),$$
the $3\times 3$ matrix is the inverse of your $B$.
So, finally 
$$X=
-\frac{251}{20}a+\frac{161}{10}b+\frac{373}{20}c.$$
