Solution using Combinatorics: Define a non-negative integer $c$ such that $a+b+c=6.$ So the problem becomes
$$
\begin{aligned}
S &= \sum_{\text{non-negative } a,b,c: \ a+b+c =6} T(a,b) \cr
\implies S &= \sum_{\text{non-negative } a,b,c: \ a+b+c =6} \binom{6}{a} \binom{6}{b} \binom{6}{6-c} \cr
\implies S &= \sum_{\text{non-negative } a,b,c: \ a+b+c =6} \binom{6}{a} \binom{6}{b} \binom{6}{c}
\end{aligned}
$$
Consider a group of $18$ people divided into three groups of $6.$ The right hand side counts the total number of ways to select $6$ people from $18$ in all the possible ways which must be equal to $ \binom{18}{6} = 18564, $ which when divided by thousand gives a remainder $564.$
Solution using Binomial Identity: This has the obvious making of Vandermonde's identity but for the inequality. So, let $c $ be a non-negative integer such that $a+b+c = 6.$ Since $a,b$ are non-negative integers, therefore $0 \le c \le 6.$
$$ T(a,b) = \binom{6}{a} \binom{6}{b} \binom{6}{6-c} = \binom{6}{a} \binom{6}{b} \binom{6}{c} \ \text{since} \left(\binom{n}{r} =\binom{n}{n-r} \right) $$
For a fixed $r,$ if $a=k,$ then $b=6-r-k$.
$$
\begin{aligned}
S &= \sum_{\text{non-negative } a,b,c: \ a+b+c =6} T(a,b) \cr
\implies S &= \sum_{r=0}^6 \sum_{k=0}^{r} \binom{6}{k} \binom{6}{r-k} \binom{6}{r} \cr
\implies S &= \sum_{r=0}^6 \binom{6}{r} \sum_{k=0}^{r} \binom{6}{k} \binom{6}{r-k}
\end{aligned}
$$
By Vandermonde's Identity, we have the following.
$$ \sum_{k=0}^{r} \binom{n}{k} \binom{m}{r-k} = \binom{n+m}{r} $$
$$ \implies S= \sum_{r=0}^6 \binom{6}{r} \binom{12}{r} = \sum_{r=0}^6 \binom{6}{r} \binom{12}{12-r} = \binom{18}{6} $$
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