I don't have complete confidence in this answer, but: Table $3$ in [Iano-Fletcher][1] lists parameters $(a_0, a_1, a_2, a_3, a_4)$ such that a general hypersurface of degree $1+\sum a_i$ in the weighted projective space  $\mathbb{P}(a_0, a_1, a_2, a_3, a_4)$ is canonically embedded and has at worst terminal isolated quotient singularities. Let $H$ be such a hypersurface and let $\pi: X \to H$ be a resolution of singularities of such a weighted hypersurface.

Then I believe that

(1) $X$ is general type.

(2) $\chi(\mathcal{O}_X) = 1-\#(i : a_i=1)$.

(3) $X$ is simply connected.

In particular, if we take a quintuple with $a_0=1$ and $a_1$, $a_2$, $a_3$, $a_4 > 1$, we get a $3$-fold with holomorphic Euler characteristic zero. Examples in Fletcher's list are $(1,2,2,3,3)$, $(1,2,3,3,5)$, $(1,2,3,4,5)$, $(1,2,2,3,9)$, $(1,3,4,5,7)$, $(1,2,3,4,11)$ and $(1,3,4,5,14)$.

<b>Regarding (1):</b> Indeed, in these examples, $H$ is canonically embedded, so $\omega = \mathcal{O}(1)$. I am a little nervous about whether I am using language correctly for a singular variety, but I believe that I am.

<b>Regarding (2):</b> The resolution $\pi : X \to H$ has $\pi_{\ast} \mathcal{O}_X = \mathcal{O}_H$ and $R^j \pi_{\ast} \mathcal{O}_X$ for $j>0$, so we can compute $H^j(H, \mathcal{O})$ instead. I did this two ways. From Theorem 7.2 in Flecther, $H^{00}(H) = \mathbb{C}$, $H^{01}(H) = H^{02}(H) = 0$ and $H^{03}(H)$ is the degree $1$ piece of the graded ring, generated by $x_0$, $x_1$, $x_2$, $x_3$, $x_4$ in degrees $(a_0, a_1, a_2, a_3, a_4)$, modulo a Jacobian ideal of relations. But the lowest degree relation in the Jacobian ideal is $\left( \sum a_i + 1 \right) - \max(a_i) > 1$, so we are just counting the number of generators in degree $1$, which is $\#(i : a_i=1)$.

I wasn't confident that I had used Theorem 7.2 correctly, so I also directly computed the Hilbert series. Put $s=\sum a_i$. The Hilbert series is
$$\frac{1-x^{s+1}}{\prod(1-x^{a_i})} = \frac{1-x^{s+1}}{\cdots+\#(i:a_i=1)x^{s-1}-x^s} = x + \#(i:a_i=1) + r(x)$$
where $r(x) \to 0$ as $x \to \infty$. When a rational function has $r(x) \to 0$ as $x \to \infty$ and denominator of the form $\prod (1-x^{a_i})$, then the coefficient of $x^n$ is quasi-polynomial in $n$ for $n$ all the way down to $0$. So we can compute $\chi(\mathcal{O}_H)$ as
$$\lim_{x \to 0} r(x) = \lim_{x \to 0} \left( \frac{1-x^{s+1}}{\prod(1-x^{a_i})} - x -  \#(i:a_i=1) \right) = 1-\#(i:a_i=1).$$

<b>Regarding (3)</b> Here is where I am a little nervous. Resolving an isolated quotient singularity shouldn't change $\pi_1$, so $\pi_1(X) = \pi_1(H)$. I want to say that I can apply the Lefschetz hyperplane theorem to the intersection of $\mathbb{P}(a_0, a_1, a_2, a_3, a_4)$ with a hyperplane in the appropriate embedding. $\mathbb{P}(a_0, a_1, a_2, a_3, a_4)$ is definitely simply connected, because all complete toric varieties are simply connected. (See EG Section 3.2 in Fulton's *Toric Varieties*.) 

But I couldn't quickly find a reference for the Lefschetz hyperplane theorem for $\pi_1$ for singular varieties which says what I want it to say. I think that this is how it works, but I'm not an expert, so I'll leave the answer with this caveat.

Thanks to [Chen Jiang's answer][2] for pointing me to Iano-Fletcher's table.


  [1]: https://homepages.warwick.ac.uk/~masda/Unpub/CPR_book/Fletcher.pdf
  [2]: https://mathoverflow.net/a/319136/297