I don't have time to write a detailed answer but the following should help.

Blast your sum with Newton's binomial formula. That gives a sum over three indices $i,j,k$. The three equations you get on these indices by fixing the monomial of interest are linearly dependent, so you are left with a sum over just one index say $i$. I believe what you get is ${}_3F_{2}$ hypergeometric sum. Your quantity of interest is also a Wigner $3jm$-symbol or Clebsch-Gordan coefficient in the literature on quantum angular momentum. In general determining the zeros of such numbers is an open problem, see my article <a href="http://link.springer.com/article/10.1007%2Fs00031-005-1111-8">"The bipartite Brill-Gordan locus and angular momentum"</a> with Jaydeep Chiplakatti for some references. If your case is sufficiently special (e.g., stretched coefficients) then you might be lucky enough to know if your coefficients are zero or not.


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As I said, I did not have time to look at your coefficient closely yesterday but from the other answers it turns out you are dealing with a trivial $3j$ zero.
In case you need more complicated analogues of your coefficients,
I am giving more details and perspective.

The Wigner $3jm$ symbol
$$
\left(\begin{array}{ccc}
j_1 & j_2 & J\\
m_1 & m_2 & m
\end{array}\right)
$$
is (up to trivial nonzero prefactors) the coefficient of $x_1^{j_1-m_1}x_2^{j_2-m_2}x_3^{J-m}$ in
$$
(x_1-x_2)^{j_1+j_2-J}(x_2-x_3)^{J+j_2-j_1}(x_3-x_1)^{J+j_1-j_2}\ .
$$

Therefore your coefficient is
$$
\left(\begin{array}{ccc}
a+c+1 & a+b+1 & b+c+1\\
0 & 0 & 0
\end{array}\right)
$$

If you look up page 2611 of the <a href="http://iopscience.iop.org/article/10.1088/0305-4470/26/11/011/meta">article</a> by Raynal, Van der Jeugt, Srinivasa Rao and Rajeswari, that we cited in my article mentioned above, you will see that your case (with all magnetic moments $m$ begin zero) is a trivial zero because the sum of the numbers on top is odd. If the sum is even the result
is nonzero and this is basically Dixon's Theorem for ${}_3F_2$ series. 

You can find many more details on $3j$, $6j$, $9j$ symbols with precise signs, prefactors and all that in my other article with Chipalkatti <a href="http://aif.cedram.org/aif-bin/item?id=AIF_2009__59_5_1671_0">"The higher transvectants are redundant"</a>.

BTW, I just looked up citations to the Raynal et al. article, and there does not seem to have been much progress on understanding the zeros of $3jm$ coefficients.