The coefficient is always zero. I use the superb conventions of Concrete Mathematics: Sums are over all integers unless otherwise indicated, and $\binom{n}{k}$ is $0$ if $k<0$ or $>n$.

Expanding by the binomial theorem, $$f_{abc} = \sum_{i,j,k} (-1)^{i+j+k} \binom{2a+1}{i} \binom{2b+1}{j} \binom{2c+1}{k} x_1^{2a+1-i+k} x_2^{2b+1-j+i} x_3^{2c+1-k+j}.$$

We want $$\begin{array}{r@{}c@{}lcr@{}c@{}l} a+c+1 &=& 2a+1-i+k &\implies& c-k &=& a-i\\ a+b+1 &=& 2b+1-j+i &\implies& a-i &=& b-j\\ b+c+1 &=& 2c+1-k+j &\implies& b-j &=& c-k \\ \end{array}$$ so $(i,j,k) = (a+r, b+r, c+r)$ for some $r$. We need to evaluate $$\sum_r (-1)^{a+b+c+3r} \binom{2a+1}{a+r} \binom{2b+1}{b+r} \binom{2c+1}{c+r}.$$

Pair off the $r$ and $1-r$ terms to get $$(-1)^{a+b+c} \sum_{r \geq 0} \left[ (-1)^{3r} \binom{2a+1}{a+r} \binom{2b+1}{b+r} \binom{2c+1}{c+r} + \right.$$ $$\phantom{(-1)^{a+b+c} \sum_{r \geq 1}} \left. (-1)^{3-3r} \binom{2a+1}{a+1-r} \binom{2b+1}{b+1-r} \binom{2c+1}{c+1-r} \right].$$

The two terms in the square brackets cancel, and the sum is zero.

The coefficient is always zero. I use the superb conventions of Concrete Mathematics: Sums are over all integers unless otherwise indicated, and $\binom{n}{k}$ is $0$ if $k<0$ or $>n \geq 0$.

Expanding by the binomial theorem, $$f_{abc} = \sum_{i,j,k} (-1)^{i+j+k} \binom{2a+1}{i} \binom{2b+1}{j} \binom{2c+1}{k} x_1^{2a+1-i+k} x_2^{2b+1-j+i} x_3^{2c+1-k+j}.$$

We want $$\begin{array}{r@{}c@{}lcr@{}c@{}l} a+c+1 &=& 2a+1-i+k &\implies& c-k &=& a-i\\ a+b+1 &=& 2b+1-j+i &\implies& a-i &=& b-j\\ b+c+1 &=& 2c+1-k+j &\implies& b-j &=& c-k \\ \end{array}$$ so $(i,j,k) = (a+r, b+r, c+r)$ for some $r$. We need to evaluate $$\sum_r (-1)^{a+b+c+3r} \binom{2a+1}{a+r} \binom{2b+1}{b+r} \binom{2c+1}{c+r}.$$

Pair off the $r$ and $1-r$ terms to get $$(-1)^{a+b+c} \sum_{r \geq 0} \left[ (-1)^{3r} \binom{2a+1}{a+r} \binom{2b+1}{b+r} \binom{2c+1}{c+r} + \right.$$ $$\phantom{(-1)^{a+b+c} \sum_{r \geq 1}} \left. (-1)^{3-3r} \binom{2a+1}{a+1-r} \binom{2b+1}{b+1-r} \binom{2c+1}{c+1-r} \right].$$

The two terms in the square brackets cancel, and the sum is zero.

The coefficient is always zero. I use the superb conventions of Concrete Mathematics: Sums are over all integers unless otherwise indicated, and $\binom{n}{k}$ is $0$ if $k<0$ or $>n$.

Expanding by the binomial theorem, $$f_{abc} = \sum_{i,j,k} (-1)^{i+j+k} \binom{2a+1}{i} \binom{2b+1}{j} \binom{2c+1}{k} x_1^{2a+1-i+k} x_2^{2b+1-j+i} x_3^{2c+1-k+j}.$$

We want $$\begin{array}{rcl} a+c+1 &=& 2a+1-i+k \\ a+b+1 &=& 2b+1-j+i \\ b+c+1 &=& 2c+1-k+j \\ \end{array}$$ so $(i,j,k) = (a+r, b+r, c+r)$ for some $r$. We need to evaluate $$\sum_r (-1)^{a+b+c+3r} \binom{2a+1}{a+r} \binom{2b+1}{b+r} \binom{2c+1}{c+r}.$$

Pair off the $r$ and $1-r$ terms to get $$(-1)^{a+b+c} \sum_{r \geq 0} \left[ (-1)^{3r} \binom{2a+1}{a+r} \binom{2b+1}{b+r} \binom{2c+1}{c+r} + \right.$$ $$\phantom{(-1)^{a+b+c} \sum_{r \geq 1}} \left. (-1)^{3-3r} \binom{2a+1}{a+1-r} \binom{2b+1}{b+1-r} \binom{2c+1}{c+1-r} \right].$$

The two terms in the square brackets cancel, and the sum is zero.

The coefficient is always zero. I use the superb conventions of Concrete Mathematics: Sums are over all integers unless otherwise indicated, and $\binom{n}{k}$ is $0$ if $k<0$ or $>n$.

Expanding by the binomial theorem, $$f_{abc} = \sum_{i,j,k} (-1)^{i+j+k} \binom{2a+1}{i} \binom{2b+1}{j} \binom{2c+1}{k} x_1^{2a+1-i+k} x_2^{2b+1-j+i} x_3^{2c+1-k+j}.$$

We want $$\begin{array}{r@{}c@{}lcr@{}c@{}l} a+c+1 &=& 2a+1-i+k &\implies& c-k &=& a-i\\ a+b+1 &=& 2b+1-j+i &\implies& a-i &=& b-j\\ b+c+1 &=& 2c+1-k+j &\implies& b-j &=& c-k \\ \end{array}$$ so $(i,j,k) = (a+r, b+r, c+r)$ for some $r$. We need to evaluate $$\sum_r (-1)^{a+b+c+3r} \binom{2a+1}{a+r} \binom{2b+1}{b+r} \binom{2c+1}{c+r}.$$

Pair off the $r$ and $1-r$ terms to get $$(-1)^{a+b+c} \sum_{r \geq 0} \left[ (-1)^{3r} \binom{2a+1}{a+r} \binom{2b+1}{b+r} \binom{2c+1}{c+r} + \right.$$ $$\phantom{(-1)^{a+b+c} \sum_{r \geq 1}} \left. (-1)^{3-3r} \binom{2a+1}{a+1-r} \binom{2b+1}{b+1-r} \binom{2c+1}{c+1-r} \right].$$

The two terms in the square brackets cancel, and the sum is zero.