If for some $k$ there exist infinitely many primes then just note $pqr|pq+rq+pr-k\implies k\geq pq+qr+pr+pqr \to \inf$ , absurd. (may be dr_civot wanted to point out that I didn't show the case if pqr\leq pq+rq+pr-k , but this is needless to since what I wrote was "infinitely many primes" or say triples,whatever result is same is pretty much obvious)

Note that $(pqr,k)=1$. Suppose contrary, that there is infinite sequence $(p_i,q_i,r_i)$, such that $p_i>q_i>r_i$ and $p_i\geq p_j$ whenever $i\geq j$. Note that $\lim_{n\to\infty} p_n=+\infty$ and since $p_i\mid q_ir_i-k$ we have $q_i^2>q_ir_i-k\geq p_i$ and hence $q_i\geq\sqrt{p_i}$ therefore $\lim_{n\to\infty}q_n=+\infty$. From $p_iq_ir_i\mid (p_iq_i-k)(q_ir_i-k)(r_ip_i-k)$ we get $p_iq_ir_i\mid p_iq_i+q_ir_i+r_ip_i-k$ and hence $p_iq_ir_i\leq p_iq_i+q_ir_i+r_ip_i-k$ or $1\leq\frac{1}{p_i}+\frac{1}{q_1}+\frac{1}{r_i}-\frac{k}{p_iq_ir_i}=L(i)$ which is contradiction since $\lim_{i\to\infty}\left(L(i)-\frac{1}{r_i}\right)=0$.

subham1729 wrote: If for some $k$ there exist infinitely many primes then just note $pqr|pq+rq+pr-k\implies k\geq pq+qr+pr+pqr \to \inf$ , absurd. Your solution is wrong because implication isn't true.

Suppose, for the sake of contradiction that there exists such $k \in \mathbb{n}$. Then, $p|qr-k,q|pr-k,r|pq-k \implies pqr|pq+qr+rp-k \implies k \le pq+qr+rp-pqr$, for infinitely many prime triplets. Then, $\frac{1}{p}+\frac{1}{q} +\frac{1}{r} > 1 $, for infinitely many triplets and so we get a contradiction.