I will provide two ways to do this. The first will be done by showing first part 3, then part 1 and then showing that part 2 follows from those. The second way will use some much more high-powered results to show something much more general, and the result will then be a special case of this.
First, we will need a little lemma that will make things easier: If $H$ is a group of order $st$ with $s$ and $t$ primes and $s > t$ then $H$ has a normal subgroup of order $s$. This follows straight from Sylow's theorems, as the number of $s$-Sylows must divide $t$ and be congruent to $1$ mod $s$ (so it is $1$ as $s>t$).
Now, we use the counting argument you already mentioned in the comments to see that $G$ will have a normal Sylow subgroup for at least one of the primes dividing its order. If it has one for $r$, we are done, so assume it has a normal $p$- or $q$-Sylow subgroup. Call this subgroup $N$.
Now, $G/N$ has order $st$ with $s$ and $t$ primes and $s>t$ ($s = r$ and either $t = p$ or $t = q$), so we can use the lemma to conclude that $G/N$ has a normal subgroup of order $s = r$).
This normal subgroup of $G/N$ then corresponds to a normal subgroup of $G$ of order either $pr$ or $qr$ (depending on whether $N$ was a $p$- or a $q$-Sylow subgroup).
But now we can use the lemma again, to conclude that this subgroup has a normal subgroup of order $r$.
To finish the proof of part 3, we note that if $P$ is a normal Sylow subgroup of $N$ and $N$ is normal in $G$, then $P$ is normal in $G$. This is because we know that since $P$ is normal in $N$, it is the only subgroup of that order in $N$, and conjugating $P$ by something from $G$ will give a subgroup of $N$ of the same order as $P$ (which must therefore be equal to $P$). If you have heard of characteristic subgroups, a simpler way to explain this is that any normal Sylow subgroup is in fact characteristic (this holds for any Hall subgroup in fact), and in general, if $M$ is characteristic in $N$ and $N$ is normal in $G$, then $M$ is normal in $G$.
Now, with part 3 proved, we can prove part 1 quite easily, since we have a normal subgroup of order $r$, call it $M$. Now, the order of $G/M$ is $pq$ and we can apply the lemma to get a normal subgroup of order $q$, which then corresponds to a normal subgroup of $G$ of order $qr$ as we wanted.
Finally, to show part 2, we now have a series of normal subgroups such that the successive quotients are abelian (in fact, the successive quotients have prime order), which shows that the group is solvable (in fact supersolvable).
The more high-powered approach is this: It is a general fact (which follows from combining Burnside's transfer theorem with the N/C theorem) that if $p$ is the smallest prime dividing the order of $G$ and $p^2$ does not divide the order of $G$ (actually, unless $p = 2$ and $3$ divides the order of the group, we just need $p^3$ to not divide the order of the group), then $G$ has a normal subgroup whose index is the order of a $p$-Sylow subgroup of $G$ (a normal $p$-complement).
We can then apply the above to see that if the order of $G$ is square free, so $|G| = p_1p_2\cdots p_n$ with $p_1>p_2>\cdots > p_n$ with all the $p_i$ primes, then $G$ has a series of normal subgroups, such that the orders are $p_1$, $p_1p_2$, $p_1p_2p_3$, and so on up to $p_1p_2\cdots p_n$ (one gets these in the reverse order by first getting one of index $p_n$, then one of index $p_{n-1}$ in that subgroup and so on). In particular, this shows that a group of square free order is supersolvable (and thus solvable).
Let $p,q,r$ be positive primes, $p<q<r$, and let $G$ be a group with $|G|=pqr$.
We know that $$n_r \equiv 1 \bmod r,\ n_r\mid pq,\ r>q,p \implies n_r \in \{1,pq\}.$$
For the worst case take $n_{r}=pq$.
By a similar argument, taking the worst case we arrive to $$n_q =r$$ $$n_p =q.$$ Here $n_i$ denotes the number of $i$-Sylow groups for $i=p,q,r$.
Now by counting we see there are $pqr-pq+1$ elements of order $r$ (with $e$), and $qr-r$ elements of order $q$. Then show that either $q$-Sylow group or $r$-Sylow group is normal. Then construct $HK$. Then the result follows.
Best Answer
I found a document that solved this problem. It's a particular case, but it helps on understanding:
http://faculty.etsu.edu/gardnerr/4127/notes/VII-37.pdf (Pages 9 and 10, Example 37.15).