Since each vector is a member of [tex]\mathbb R^3[/tex], the vectors will span [tex]\mathbb R^3[/tex] if they form a basis for [tex]\mathbb R^3[/tex], which requires that they be linearly independent of one another.
To show this, you have to establish that the only linear combination of the three vectors [tex]c_1\mathbf v_1+c_2\mathbf v_2+c_3\mathbf v_3[/tex] that gives the zero vector [tex]\mathbf0[/tex] occurs for scalars [tex]c_1=c_2=c_3=0[/tex].
[tex]c_1\begin{bmatrix}3\\2\\1\end{bmatrix}+c_2\begin{bmatrix}-1\\-2\\-4\end{bmatrix}+c_3\begin{bmatrix}1\\1\\-1\end{bmatrix}=(0,0,0)\iff\begin{bmatrix}3&-1&1\\2&-2&1\\1&-4&-1\end{bmatrix}\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}[/tex]
Solving this, you'll find that [tex]c_1=c_2=c_3=0[/tex], so the vectors are indeed linearly independent, thus forming a basis for [tex]\mathbb R^3[/tex] and therefore they must span [tex]\mathbb R^3[/tex].
