$B= [a,b,c]$ 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
$$\begin{bmatrix} -3&-3&1\\ 2&2&6\\ 1&-1&1\\ \end{bmatrix}$$
On
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 $$X= -\frac{251}{20}a+\frac{161}{10}b+\frac{373}{20}c.$$
Hint: You have to check if your vectors are linearly independent. This is sufficient as you have 3 vectors for a 3D space. For a 3D-System this is equivalent to $det(A)\neq 0$. Where A is the matrix that you constructed.