= Solution
A <Hall subgroup> for a set of primes $\pi$ is a <subgroup> $H\leq G$ such that every prime divisor of $|H|$ belongs to $\pi$, while no prime divisor of $[G:H]$ belongs to $\pi$. Equivalently $|H|=|G|_\pi$, the product of the full prime powers in $|G|$ belonging to $\pi$.
The <Hall conjugacy and embedding in finite soluble groups> theorem says that, for every $\pi$, a finite <soluble group> has a Hall $\pi$-subgroup, any two such <subgroups> are conjugate, and every $\pi$-subgroup is contained in one. We prove all three assertions together by induction, including the needed complement argument rather than assuming it.
First establish <coprime splitting over an elementary abelian normal subgroup>. Suppose $V\trianglelefteq E$ is an elementary <Abelian> $r$-group and $D=E/V$ has order $m$ prime to $r$. Write $V$ additively. <Conjugation> defines a well-defined action of $D$ on $V$ because $V$ is <Abelian>. Choose a section $s:D\to E$, with $s(1)=1$, and let its <extension cocycle> be defined by $s(x)s(y)=f(x,y)s(xy)$. Associativity gives
$$
f(x,y)+f(xy,z)=x f(y,z)+f(x,yz).
$$
Sum over $z\in D$. With $S(x)=\sum_z f(x,z)$ this becomes
$$
mf(x,y)+S(xy)=xS(y)+S(x).
$$
Multiplication by $m$ is invertible on $V$, so setting $b(x)=m^{-1}S(x)$ gives $f(x,y)=b(x)+xb(y)-b(xy)$. The new section $s'(x)=(-b(x))s(x)$ has zero cocycle and is a homomorphism. Its image is a complement $K$ to $V$, proving $E=V\rtimes K$.
Every other complement has the form $\{h(x)s'(x):x\in D\}$, with $h(xy)=h(x)+xh(y)$. Summing this identity over $y$ and putting $T=\sum_y h(y)$ gives $mh(x)=T-xT$. Hence $h(x)=v-xv$, where $v=m^{-1}T$, and the other complement is $vKv^{-1}$. The same argument applies on any <subgroup> $D_0\leq D$: a <subgroup> intersecting $V$ trivially is a graph of such a one-cocycle over its image $D_0$, so <conjugation> by an element of $V$ carries it into $K$. This proves the complement-conjugacy and embedding assertions needed below.
Now let $G\ne1$ be finite soluble, and choose a <minimal normal subgroup> $V$. It is elementary <Abelian> of some prime characteristic $r$, by the minimal-normal argument in the series discussion. Induction gives Hall <subgroups> and the embedding and conjugacy assertions in $G/V$. Let $\overline H$ be a Hall $\pi$-subgroup there and $M$ its full preimage.
If $r\in\pi$, then $M$ itself is a Hall $\pi$-subgroup of $G$. Every Hall $\pi$-subgroup $H$ contains $V$, since $HV$ is a $\pi$-subgroup and $H$ already has the largest possible $\pi$-order. Thus $H$ is the full preimage of its image in $G/V$, and quotient conjugacy lifts to conjugacy in $G$. Given any $\pi$-subgroup $U$, induction puts its quotient image into some $\overline H$, whose preimage $M$ contains $U$.
If $r\notin\pi$, then $M/V=\overline H$ has order prime to $r$. The complement construction produces $K\leq M$ of order $|\overline H|=|G|_\pi$, hence a Hall <subgroup> of $G$. Any Hall <subgroup> has trivial intersection with $V$ and a Hall image in $G/V$. Conjugating two Hall <subgroups> first aligns these images, after which both are complements to $V$ in the same $M$; the complement-conjugacy argument then conjugates them by an element of $V$. For an arbitrary $\pi$-subgroup $U$, quotient induction puts its image into some $\overline H$. Then $U\subseteq M$ and $U\cap V=1$, so the embedding argument conjugates $U$ into the chosen complement $K$. Equivalently $U$ lies in a conjugate of $K$. This completes the induction and all three parts of the <Hall theorem for soluble groups>.
For the specific <general linear group over a finite field>, counting ordered bases gives
$$
|GL_3(2)|=(8-1)(8-2)(8-4)=168.
$$
There are seven lines and seven planes in $\mathbb F_2^3$. The <group> is transitive on each family, so line and plane stabilizers both have order $24$ and index seven. To show that these exhaust the index-seven <subgroups>, let $H$ have order $24$. It contains a Sylow $2$-subgroup of order eight, which can be conjugated to
$$
U=\left\{\begin{pmatrix}1&a&b\\0&1&d\\0&0&1\end{pmatrix}:a,b,d\in\mathbb F_2\right\}.
$$
The <group orbits> of $U$ on the seven nonzero <vectors> are
$$
\{e_1\},\quad\{e_2,e_1+e_2\},\quad
\{e_3,e_1+e_3,e_2+e_3,e_1+e_2+e_3\},
$$
of sizes $1,2,4$. An $H$-orbit is a union of these. The <group> $H$ cannot be transitive on seven points, since seven does not divide $24$. Therefore it has an orbit of size at most three. A size-one orbit fixes $e_1$. A size-two orbit must be $\{e_2,e_1+e_2\}$, whose <vector> sum $e_1$ is consequently fixed. A size-three orbit must be the union of the first two $U$-orbits, the nonzero part of the plane $\langle e_1,e_2\rangle$, so that plane is invariant. In every case $H$ lies in a line or plane stabilizer, and equality follows from their equal orders.
The two stabilizer families are not conjugate. A line stabilizer fixes its nonzero <vector> and is transitive on the other six nonzero <vectors>: extend that fixed <vector> and either chosen other <vector> to a basis, and map one such basis to the other. Thus it has no invariant three-element subset of nonzero <vectors> and fixes no plane. In contrast a plane stabilizer fixes its defining plane. <Conjugation> sends line stabilizers to line stabilizers, so cannot send one to a plane stabilizer. We obtain
$$
\boxed{\text{Exactly two conjugacy classes of index-seven subgroups: line stabilizers and plane stabilizers.}}
$$
They are <Hall subgroups> for $\pi=\{2,3\}$ but are not conjugate. This explicitly shows why the solubility hypothesis of the <Hall theorem for soluble groups> matters.
Finally, an index-three <subgroup> would give a nontrivial transitive <coset> homomorphism $GL_3(2)\to S_3$. Its kernel is normal; the permitted simplicity of $GL_3(2)$ forces that kernel to be trivial, because the action is nontrivial. But an injective map from a <group> of order $168$ into one of order six is impossible. Hence \b[$GL_3(2)$ has no <subgroup> of index three].
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