= Solution
A family $x_1,\ldots,x_n$ is <linearly independent> if $\sum_{i=1}^n a_ix_i=0$ forces $a_1=\cdots=a_n=0$. For the three specified vectors, direct substitution gives
$$
\boxed{4x_1-3x_2-2x_3=0.}
$$
The coefficients are not all zero, so \b[the vectors are linearly dependent].
For the implication question, using the usual interpretation that a <basis> contains the listed family as independent entries, the answers in order (i)–(x) are
$$
\boxed{\mathrm T,\ \mathrm T,\ \mathrm F,\ \mathrm F,\ \mathrm T,\ \mathrm F,\ \mathrm T,\ \mathrm T,\ \mathrm T,\ \mathrm T.}
$$
The key distinctions are that the unrestricted zero combination always exists, whereas <linear dependence> requires a nontrivial combination; and dependence does not necessarily let the specified third vector be solved in terms of the first two. For example $x=y=e_1$, $z=e_2$ makes (iii) and (iv) false, while $x=e_1,y=e_2,z=e_3$ disproves (vi).
There is a set-versus-family qualification in (v). If “contains” is interpreted as literal set containment, repeated nonzero vectors need only occur once in the <basis>. Then $x=y=e_1,z=e_2$ is dependent as a family but all three belong to the basis $\{e_1,e_2\}$, so (v) becomes false. With three distinct vectors, or with family-containment as above, <basis extension> gives (v) true. Implication (x) remains true on either reading.
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