Only two co-efficient

Let roots are $r_i$ where $i \leq i \leq 6$

From the info given in the question $\displaystyle \sum_{i=1}^{6} r_i=12$ and $\displaystyle \prod_{i=1}^{6} r_i=64$ .

As the roots are positve reals using A.M.-G.M. inequality we get

$\dfrac{r_1+r_2+....+r_6}{6} \geq \sqrt[6]{r_! r_2...r_6} \\ \implies \displaystyle \sum_{i=1}^{6} r_i \geq \displaystyle \prod_{i=1}^{6} r_i$

So, equality is satisfying.Therefore $r_1=r_2=....=r_6$

So, value of each root is $\dfrac{12}{6}=2$

Now we can express $f(x)$ becomes $f(x)=(x-2)^6$

$f(1)=1;f(2)=0;f(3)=1$