= Solution
\b[No.] Every one of the three <vectors> has equal first and second <Cartesian coordinates>, so their <linear span> is contained in the <plane> $x=y$. That <linear subspace> has <dimension> two and cannot equal $\mathbb R^3$.
More explicitly, call the three given <vectors> $\mathbf u,\mathbf v,\mathbf w$ in their listed order. Then
$$
3\mathbf u+\mathbf v-2\sqrt3\,\mathbf w=0.
$$
This nontrivial <linear relation> proves <linear dependence>, whereas a <basis> must be <linearly independent>. Being <unit vectors> alone does not ensure that they form a <basis>.
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