Hint: $11+3<2017$

We use a method called Mixing Variables.

Let $f(x,y,z)= 11xy+3xz+2017yz$ . We have

$f(0,\frac{x}{2}+y,\frac{x}{2}+z)= 2017(\frac{x}{2}+y)(\frac{x}{2}+z)$

$\geq \frac{2017xy}{2} + \frac{2017xz}{2} + 2017yz \geq f(x,y,z)$ .

What this means is that it is always best to take $x=0$ for maximising $f(x,y,z)$ .

Also, $f(0,\frac{x}{2}+y,\frac{x}{2}+z)= 2017(\frac{x}{2}+y)(\frac{x}{2}+z)$

$\leq 2017(\frac{(\frac{x}{2}+y)+ (\frac{x}{2}+z)}{2})^{2} = 2017(\frac{1000}{2})^{2}$

$=2017\times (500)^2$

Therefore $f(x,y,z) \leq \boxed{504250000}$ .

Equality holds iff $x=0,y=z=500$ .