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 a ${}_3F_{2}$ terminating hypergeometric sum. Your quantity of interest is also a Wigner $3jm$-symbol or Clebsch-Gordan coefficient from the literature on quantum angular momentum. In general, determining the zeros of such numbers is an open problem, see my article "The bipartite Brill-Gordan locus and angular momentum" 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.

Edit:
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 article 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 "The higher transvectants are redundant".

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.