How to find out if the three $3\times1$ vectors are a basis for $\mathbb R^3$ 
Is $\begin{bmatrix}1\\1\\-2\end{bmatrix}, \begin{bmatrix}7\\0\\-5\end{bmatrix}$ and $\begin{bmatrix}-5\\-1\\2\end{bmatrix}$ a basis for $\mathbb{R}^3$?

I know that they are not linearly independent because they vectors are not multiples of each other. With that being said, since they are not linearly independent, I do not know how to further test to see if they are a basis for $\mathbb{R}^3$ 
 A: Definition: A set of vectors $\{\mathbf v_1, \dots, \mathbf v_k\}\subset \Bbb R^n$ is linearly independent if $$\lambda_1\mathbf v_1 + \cdots + \lambda_k\mathbf v_k = \mathbf 0 \implies \lambda_1 = \cdots = \lambda_k=0$$
where $\lambda_1, \dots, \lambda_k \in \Bbb R$.  If a set of vectors is not linearly independent, it is linearly dependent.
Definition: A set $\{\mathbf v_1, \dots, \mathbf v_k\}\subset \Bbb R^n$ is a basis if it is a linearly independent spanning set.

So for your set of vectors, we know there are the right number for a basis (because the dimension of $\Bbb R^3$ is $3$), so we just need to see if they are linearly independent.
$$\lambda_1 (1,1,-2) + \lambda_2 (7,0,-5) + \lambda_3 (-5,-1,2) = (0,0,0) \iff \pmatrix{1 & 7 & -5 \\ 1 & 0 & -1 \\ -2 & -5 & 2}\pmatrix{\lambda_1 \\ \lambda_2 \\ \lambda_3} = \pmatrix{0 \\ 0 \\ 0}$$
Solve this system of equations and if the only solution is $\lambda_1=\lambda_2=\lambda_3=0$ then your set is a linearly independent set.  If there are any other solutions then your set is linearly dependent and thus not a basis.
