= Solution
Any nonzero <vector> in $\mathbb F_2^3$ extends to a <basis>. Sending one such <basis> to another gives an element of the <general linear group>, so its <group action> on the seven nonzero <vectors> is a <transitive group action>. Counting successive independent columns gives
$$
|G|=(8-1)(8-2)(8-4)=168.
$$
By the <orbit-stabilizer theorem>, the <stabilizer> $H$ of a chosen nonzero <vector> $v$ has order $168/7=24$.
There are three two-dimensional <linear subspaces> containing $v$, corresponding to the three one-dimensional <linear subspaces> of $\mathbb F_2^3/\langle v\rangle$. The <stabilizer> $H$ acts transitively on them: extend a <basis> of either plane beginning with $v$ to a <basis> of the whole <vector space>. Thus the <stabilizer> of a flag $\langle v\rangle<P$ has order $24/3=8$.
If $Q\le G$ has order eight, each <orbit of a group action> of $Q$ on the seven nonzero <vectors> has size a power of two. Since seven is odd, at least one <orbit of a group action> has size one. Thus $Q$ fixes a nonzero <vector> $v$. Its <group action> on the three planes containing $v$ likewise has a fixed plane $P$. Therefore $Q$ is contained in, and by order equals, the <stabilizer> of that flag. The <group action> of $G$ is transitive on these flags, so all such <subgroups> are <conjugate subgroups>.
Distinct flags give distinct <stabilizers>. In a <basis> adapted to a flag, its order-eight <stabilizer> consists of
$$
\begin{pmatrix}1&a&b\\0&1&c\\0&0&1\end{pmatrix},\qquad a,b,c\in\mathbb F_2.
$$
Its only common fixed nonzero <vector> is $e_1$. Its only invariant plane is $\langle e_1,e_2\rangle$: an invariant plane is the <kernel of a linear map> of a nonzero <linear functional>, and over $\mathbb F_2$ that <linear functional> must be fixed; the displayed <matrices> have only the fixed nonzero <linear functional> $e_3^*$. Hence each <subgroup> specifies its flag uniquely. There are seven choices of $v$ and three choices of $P$, giving
$$
\boxed{21\text{ subgroups of order }8.}
$$
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