= Solution
The converted source omits $r$ from the product; the PDF asks about $N=pqr$. Fix the odd prime $r$, and suppose $pqr$ is Carmichael. Its three prime factors are distinct by the <Korselt criterion>. Relabel $p,q$ so that $p<q$.
The divisibility conditions include
$$
q-1\mid pr-1,\qquad p-1\mid qr-1.
$$
Write
$$
pr-1=b(q-1)
$$
for a positive integer $b$. Since $p<q$,
$$
pr-1<r(q-1),
$$
so $1\leq b<r$. Moreover,
$$
q=\frac{pr+b-1}{b}.
$$
Multiplying the second divisibility by $b$ and reducing modulo $p-1$ gives
$$
p-1\mid b(qr-1)
=pr^2+r(b-1)-b
\equiv(r-1)(r+b)\pmod{p-1}.
$$
The positive integer on the right is at most $(r-1)(2r-1)$. Consequently
$$
p\leq(r-1)(2r-1)+1.
$$
There are only finitely many possible primes $p$ and integers $b\in\{1,\ldots,r-1\}$, and each pair determines at most one $q=(pr+b-1)/b$. Therefore
$$
\boxed{\text{for fixed }r,\text{ only finitely many prime pairs }(p,q)
\text{ make }pqr\text{ Carmichael}.}
$$
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