Problem: Are there any triples $(a,b,c)$ of positive integers such that
$(a-2)(b-2)(c-2)+12$ is a prime number
that properly divides the positive number
$a^2+b^2+c^2+abc-2017$?

Solution: No such $(a,b,c)$.

Assume not.
Let $x=a-2$, $y=b-2$, $z=c-2$, hence $x,y,z \ge -1$.
\begin{align*}
  a^2+b^2+c^2+abc-2017
  &= (x+2)^2 + (y+2)^2 + (z+2)^2 \\ &
  + (x+2)(y+2)(z+2) - 2017 \\
  &= (x+y+z+4)^2 + (xyz+12) - 45^2.
\end{align*}
Thus the divisibility relation becomes
\[ p = xyz+12 \mid \left( x+y+z+4 \right)^2 - 45^2 > 0 \]
so either
\begin{align*}
  p &= xyz+12 \mid x+y+z-41 \\
  p &= xyz+12 \mid x+y+z+49
\end{align*}

Assume $x \ge y \ge z$, hence $x \ge 14$ (since $x+y+z \ge 41$).
We now eliminate several edge cases
to get $x,y,z \neq -1$ and a little more:
\begin{claim*}
  We have $x \ge 17$, $y \ge 5$, $z \ge 1$, and $\gcd(xyz, 6) = 1$.
\end{claim*}
\begin{proof}
  First, we check that neither $y$ nor $z$ is negative.
  \begin{itemize}
  \ii If $x > 0$ and $y=z=-1$, then we want $p=x+12$
  to divide either $x-43$ or $x+47$.
  We would have $0 \equiv x-43 \equiv -55 \pmod p$
  or $0\equiv x+47 \equiv 35 \pmod p$,
  but $p > 11$ contradiction.

  \ii If $x, y > 0$, and $z = -1$, then $p = 12-xy > 0$.
  However, this is clearly incompatible with $x \ge 14$.
  \end{itemize}
  Finally, obviously $xyz \neq 0$ (else $p=12$).
  So $p = xyz + 12 \ge 14 \cdot 1^2 +12 = 26$ or $p \ge 29$.
  Thus $\gcd(6,p) = 1$ hence $\gcd(6,xyz)=1$.

  We finally check that $y=1$ is impossible, which forces $y \ge 5$.
  If $y=1$ and hence $z=1$ then $p=x+12$ should divide either
  $x+51$ or $x-39$.
  These give $39 \equiv 0 \pmod p$ or $25 \equiv 0 \pmod p$,
  but we are supposed to have $p \ge 29$.
\end{proof}

In that situation $x+y+z-41$ and $x+y+z+49$ are both even,
so whichever one is divisible by $p$
is actually divisible by $2p$.
Now we deduce that:
\[ x+y+z+49 \ge 2p = 2xyz + 24 \implies 25 \ge 2xyz-x-y-z. \]
But $x \ge 17$ and $y \ge 5$ thus
\begin{align*}
  2xyz-x-y-z &= z(2xy-1) - x - y \\
  &\ge 2xy - 1 - x - y \\
  &> (x-1)(y-1) > 60
\end{align*}
which is a contradiction.
Having exhausted all the cases we conclude no solutions exist.

\begin{remark*}
  The condition that $x+y+z-41 > 0$ (which comes from ``properly divides'')
  cannot be dropped. Examples of solutions in which $x+y+z-41 = 0$
  include $(x,y,z) = (31,5,5)$ and $(x,y,z) = (29,11,1)$.
\end{remark*}