The answer is “No”.

We want to see if \(\alpha = \beta=\gamma=0\) is the only solution to \[\alpha \begin{bmatrix} 1\\0\\1\end{bmatrix}+ \beta\begin{bmatrix} 1\\-1\\1\end{bmatrix}+ \gamma\begin{bmatrix} 1\\1\\1\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}.\] By comparing the two sides, we obtain the system \begin{eqnarray*} \alpha + \beta + \gamma & = & 0\\ -\beta + \gamma & = & 0 \\ \alpha + \beta + \gamma & = & 0 \end{eqnarray*} We see that the third equation is superfluous. Setting \(\gamma = 1\), we obtain \(\beta = 1\) and \(\alpha = -2\). In other words, \(-2 \begin{bmatrix} 1\\0\\1\end{bmatrix} +\begin{bmatrix} 1\\-1\\1\end{bmatrix}+ \begin{bmatrix} 1\\1\\1\end{bmatrix} = \begin{bmatrix} 0 \\ 0\\ 0\end{bmatrix}\).