Put the proposed vectors into the columns of a matrix $A$. They span $\mathbb R^3$ iff $A$ has rank three, equivalently a pivot in every row. For columns $(1,0,0)$, $(0,1,0)$, $(1,1,1)$, the determinant is $1$, so they span. Three vectors confined to the plane $z=0$ cannot span $\mathbb R^3$.
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