= Solution
{c}
Subtracting the first <vector> from each of the other two preserves the <basis> property by <1/a/iii>. The resulting family is
$$
e_1,\quad(0,\alpha,\alpha^2)^T,\quad(0,\beta,\beta^2)^T.
$$
Its last two <vectors> belong to the plane with first coordinate zero. By <1/b>, they are <linearly independent> there precisely when
$$
\alpha\beta^2-\beta\alpha^2=\alpha\beta(\beta-\alpha)\ne0.
$$
In that case their two-dimensional <span> is the whole plane, and adjoining $e_1$ gives a <basis> of $\mathbb R^3$. If the displayed expression vanishes, a nontrivial <linear relation> among those two <vectors> prevents a <basis>. Thus
$$
\boxed{\alpha\ne0,\quad\beta\ne0,\quad\alpha\ne\beta.}
$$
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