$h(z)=|z-a| \cdot |z-b| \cdot |z-c|$, max value of $h$ is attained 
Let $a,b,c$ be non-collinear points in complex plane, $\Delta$ be the closed triangular region of the plane with vertices $a,b,c$. for $z\in\Delta$, let $$h(z)=|z-a| \cdot|z-b| \cdot |z-c|$$
Then max value of $h$
1. is not attained at any point of $\Delta$
2. is attained at an interior point of $\Delta$
3. is attained at the centre of gravity of $\Delta$
4. is attained at the boundary point of $\Delta$

4 is correct due to Maximum Modulus Principle. Just confirm me please.
 A: Yes, by the maximum principle applied to $(z-a)(z-b)(z-c)$, the maximum of $h$ is attained at a (not the) boundary point of the triangle. 
The specific triangular shape does not matter; the same holds for any domain. 

I'll add a similar, somewhat entertaining example: distribute $n$ points $z_1,\dots,z_n$ uniformly along the unit circle. The maximum of the product $\prod_{k=1}^n |z-z_k|$ over the disk $|z|\le 1$ is equal to $2$ (and is attained at the boundary). It's not exactly intuitive that the maximum is independent of $n$. 
A: Except your question, we can also answer to the maximum value of the function $h(x)$.
Let $f(z)=(z-a)(z-b)(z-c), \; \forall z\in\Delta.$ Then $f$ agrees to Maximum Modulus Principle, hence $$|f(z)|\leq\max\{|f(z)|\vert z\in\partial\Delta\},\; \forall z\in Int\Delta$$ i.e. $\max\{h(z)\vert z\in\Delta\}=\max\{h(z)\vert z\in\partial\Delta\}$ where $h(z)=|f(z)|.$
Since $f(a)=f(b)=f(c)=0,$ the maximum occurs at $z\in(a,b)\cup(b,c)\cup(c,a).$ Since $\Delta$ is an equilateral triamgle, wlog let $z\in(a,b).$ Then $$z=\lambda a+(1-\lambda )b,\; \lambda\in(0,1).$$
On that case, $h(z)=\lambda(1-\lambda)|a-b|^2|\lambda a+(1-\lambda )b-c|.$ Let $|a-b|=|b-c|=|c-a|=p.$ Then $h(z)=\lambda(1-\lambda)p^2|\lambda a+(1-\lambda )b-c|.$
If you look at the triangle $\Delta,$ from the fact that the height of the triangle is $\frac{p\sqrt{3}}{2},$ using the Pythagorean Theorem, we see that $|z-c|=|\lambda a+(1-\lambda )b-c|=p\sqrt{\lambda^2-\lambda+1},$ so $h(z)=\lambda(1-\lambda)p^3\sqrt{\lambda^2-\lambda+1}.$
We see that $\max h(z)=\max\{\lambda(1-\lambda)p^3\sqrt{\lambda^2-\lambda+1}|\lambda\in(0,1)\}=\frac{\sqrt{3}}{2}p^3.$
