Is vector $w$ in the span of $\{v_1,v_2,v_3\}$? Let $v_1 = (1,0,-1)^T$, $v_2 = (2, 1, 3)^T$, $v_3 = (4,2,6)^T$ and $w = (7,4,7)^T$

From what I understand, if I set up an augment matrix in the form 

\begin{pmatrix}1&amp;2&amp;4&amp;7\\
0&amp;1&amp;2&amp;4\\
-1&amp;3&amp;6&amp;7\\
\end{pmatrix}

And reduce it using Gaussian elimination, if it is consistent, $w$ is in fact in the span of $\{v_1,v_2,v_3\}$. Else, if it is inconsistent, $w$ does not span $\{v_1,v_2,v_3\}$

This, $w$ does not span $\{v_1,v_2,v_3\}$. Is this true?
You are correct. If $w$ were an element of $\operatorname{span}(v_1, v_2, v_3)$, then we could represent $w$ as $w=xv_1 + yv_2 + zv_3$.  Thus there'd be a solution to the matrix equation $$\pmatrix{1&amp;2&amp;4\\0&amp;1&amp;2\\-1&amp;3&amp;6}\pmatrix{x \\ y \\ z} = \pmatrix{7 \\ 4 \\ 7}$$

Solving this, we see that
$$RREF\left(\begin{array}{ccc|c}1&amp;2&amp;4&amp;7\\
0&amp;1&amp;2&amp;4\\
-1&amp;3&amp;6&amp;7
\end{array}\right) = \left(\begin{array}{ccc|c}1 &amp; 0 &amp; 0 &amp; 0\\ 0 &amp; 1 &amp; 2 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1\end{array}\right) \\ \iff \begin{cases} x=0 \\ y+2z=0 \\ 0=1\end{cases}$$ but that last equation is never true so this system of equations has no solution.

Thus $w \not\in \operatorname{span}(v_1, v_2, v_3)$.

In other words, @tommy is wrong: $w$ is NOT a linear combination of those $3$ vectors.
yes as it is a linear combination of those 3 vectors