$pq-6$, $qr-6$ and $rp-6$ are all perfect squares. Prove that $p+q+r-9$ is also a perfect square. 
$p$, $q$ and $r$ are distinct prime numbers such that $pq-6$, $qr-6$ and $rp-6$
  are all perfect squares. Prove that $p+q+r-9$ is also a perfect
  square.

Now comes a part where I explain what I have tried and concluded so far. In short: nothing, except that only one prime number can be 2. In fact, finding a working example proved to be a challenge. And this is not homework, I'm too old for it.
EDIT: Actually, I turned out to be smarter than I thought. Suppose that we have three odd primes $p,q,r$. In that case at least two are equal to 1 or 3 (modulo 4). In both cases their product is equal to 1 (modulo 4). Subtract 6 and and you get 3 (modulo 4). And such number cannot be a perfect square. 
So at least one prime number has to be $2$, say $p$. I solved 1/3 of the problem. Probably enough to avoid downvotes :) 
 A: By the work you've carried out in the edit, we may assume without loss of generality that $p=2$ and $p<q<r$. This transforms our statements to the following: given primes $q<r$ and  that $2q-6,2r-6,qr-6$ are squares, show that $q+r-7$ is also a square.
Conjecture: if $2q-6=s^2$ and $2r-6=t^2$, then $t=s+2$ and $q+r-7=(s+1)^2$. Further, $\sqrt{qr-6} = \frac{q+r}{2}-1$. This seems to match with the results of Will Jagy in another answer in the thread. Unfortunately, proving this has eluded me for a while and now I must go do other things. I'll come back to this.
A: Not a complete answer but a start:
A perfect square must equal $0$ or $1$ modulo $4$. Thus, $6+n^2$ must be $2$ or $3$ modulo $4$. These remainders after division by $4$ can be built in the following ways:
$$1\times 2\equiv 2$$
$$2\times 3\equiv 2$$
$$1\times 3\equiv 3$$
In other words, remainders of both factors must not be the same! The only option is thus, that $p,q,r$ must have remainders modulo $4$ equal to $1$, $2$ and $3$ (must be different). That in turn means, that one of them must be $q=2$. It follows that $p+q+r-9\equiv 1 \mod 4$ - a perfect square of an odd number.
Now that we know a bit more about what $p,q,r$ must be, it might be easier to proceed. Especially the fact that one of them is $2$, should help simplify matters.
The original statement is now:
$$p=3+2n^2$$
$$r=3+2l^2$$
$$pr=6+m^2$$
and
$$p+r-7=x^2$$
where $p$ and $r$ are odd primes, $p=1\mod 4$, $r=3\mod 4$, and $x$ is an odd number. $n$ and $m$ must be odd and $l$ must be even, and $p+r$ turns out to be divisible by $8$.
At the end, I suspect we will need the fact that $p$ and $r$ are primes - just modular arithmetics won't be enough. Unique factorization will play a role.

Now observe the same equations modulo $3$. Perfect squares can only be equal to $0$ or $1$ modulo $3$. The $0$ case is only if they are divisible by $3$. It can easily be shown that $n$ and $l$ are not divisible by $3$ except for the trivial solution $(p,q,r)=(3,2,5)$. Thus we have
$$p,r\equiv 2 \mod 3$$
$$p-r=2(n-l)(n+l)\equiv 0\mod 6$$
