= Solution
First establish the <binomial theorem> needed throughout this question. For integers $0\leq m\leq n$, define the <binomial coefficient> $\binom nm$ to be the number of $m$-element <subsets> of an $n$-element set. Counting ordered selections and then forgetting their order gives
$$
\binom nm=\frac{n!}{m!(n-m)!},
$$
with the <factorial> convention $0!=1$. In the product of $n$ factors $(1+z)$, choose $z$ from exactly $m$ factors and $1$ from the others. Each choice contributes $z^m$, and there are $\binom nm$ choices. The distributive law therefore proves
$$
(1+z)^n=\sum_{m=0}^n\binom nm z^m
$$
for every <complex number> $z$, including $n=0$ by the empty-product convention.
To evaluate the even-indexed sum, use the <even part of a polynomial>: averaging the <binomial theorem> at $z$ and $-z$ cancels the odd powers. For $z=i\sqrt3$,
$$
\sum_{k=0}^{3n}(-3)^k\binom{6n}{2k}
=\frac{(1+i\sqrt3)^{6n}+(1-i\sqrt3)^{6n}}2.
$$
The two <complex numbers> satisfy $1\pm i\sqrt3=2e^{\pm i\pi/3}$. Their $6n$th powers are both $2^{6n}$, since $e^{\pm2\pi in}=1$. Thus \b[the sum is]
$$
\boxed{2^{6n}}.
$$
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