= Solution
Let
$$
F(n)=(2n+1)(3n+1)(5n+1).
$$
For every prime $p>5$, the congruence $F(n)\equiv0\pmod p$ excludes the three distinct residue classes
$$
n\equiv-2^{-1},-3^{-1},-5^{-1}\pmod p.
$$
The finitely many smaller primes only alter the implied constant. The <dimension-three upper-bound sieve>, used with $z=x^{1/2}$, therefore gives
$$
|\{n\leq x:\gcd(F(n),P(z))=1\}|
\ll x\prod_{5<p\leq z}\left(1-\frac3p\right)
\ll\frac{x}{(\log z)^3}
\ll\frac{x}{(\log x)^3},
$$
where the middle estimate follows from <Mertens theorem>.
If all three linear forms are prime, then either $F(n)$ has no prime divisor at most $z$, or one of the three forms itself equals such a prime. The latter possibility contributes only $O(\pi(z))=O(x^{1/2})$, which is absorbed by $x/(\log x)^3$. Hence the required number of $n$ is $\ll x/(\log x)^3$.
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