From $a\mid Nbc + b + c$ we have $x\mid Ndyz + y + z$. If $\gcd(x,y) = m$, then $m\mid z$, so $m = 1$, hence $x,y,z$ are pairwise coprime (because $x\mid Ndyz + y + z$, $y\mid Ndzx + z + x$, $z\mid Ndxy + x + y$). Consider $T = Ndyz + Ndzx + Ndxy + x + y + z$; then $x\mid T$, $y\mid T$, $z\mid T$, so $xyz\mid T$. Suppose $\gcd(d,xyz) = r$; then $r\mid T=Ndyz + Ndzx + Ndxy + x + y + z$, so $r\mid x + y + z$, and we're done.

But the problem requires "if and only if". It only has been proved that, if $N$ exists satisfying the former divisibility conditions, then the latter conditions on $a,b,c$ are also satisfied. It remains to prove that, under those latter conditions, some $N$ exists satisfying the former divisibility conditions.

If $\gcd (d,xyz)\mid x + y + z$ and $x,y,z$ are pairwise corprime, then $\gcd \left( {xy + yz + zx,xyz} \right) = 1$. By Bézout's relation there exist $N,k$ satisfying $kxyz - Nd(xy + yz + zx) = x + y + z$, so $xyz\mid x + y + z + Nd(xy + yz + zx)$, hence we have $x\mid Ndyz + y + z$, $y\mid Ndzx + z + x$, $z\mid Ndxy + x + y$, and we're done.