There are $12^{12}$ = 8,916,100,448,256 total possible outcomes. I decided to approach it by calculating the probability of rolling the $12$ dice WITHOUT getting a triple and subtract from $1$. I figured the first two dice could be the same and each additional die would be one less:
$12\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2 = 5,748,019,200$. So:
$1-\dfrac{12 \cdot 12!}{12^{12}} = 99.936%$ probability of getting a triple
But when I plug the data into Wolfram Alpha, they give me a probability of slightly over 50%. What gives? I'm clearly approaching this completely wrong. Thanks!
To clarify, I am trying to find the probability of rolling exactly one triple with the other 9 numbers different from each other and from the triple.
In principle, you can definitely calculate the probability of not getting any triples and subtract that from $1$. This is tricker than you're accounting for, though, because you may have up to six pairs, not just one. If you have $1\le k \le 6$ pairs, you may choose their values in ${{12}\choose{k}}$ ways; there are then $12-2k$ dice left over, which must be singletons, with values that may be chosen in ${{12-k}\choose{12-2k}}={{12-k}\choose{k}}$ ways. Finally, there are $12!$ ways to order the dice, of which $12!/2^k$ are distinct. Summing up, $$ p_{\text{no triples}}=\frac{12!}{12^{12}}\sum_{k=1}^{6}\frac{{{12}\choose{k}}{{12-k}\choose{k}}}{2^k}=\frac{(12!)^2}{12^{12}}\sum_{k=1}^{6}\frac{1}{(k!)^2(12-2k)!2^k}\approx 0.3126. $$ This is assuming that four-or-more of a kind also counts as containing a triple. If not (i.e., if you want the probability of not having exactly $3$ of any value), you have additional configurations to consider, making this probability slightly higher and the calculation even more complex.
Update:
(Some years later.) OP's clarification, which was posted after my answer, was that they want the probability of getting exactly one triple and nine singletons. To get this, choose the value of the triple (in $12$ ways), and which rolls constitute the triple (in ${{12}\choose{3}}=\frac{12\cdot11\cdot10}{3!}=220$ ways), and the values of the remaining rolls in $11\times10 \times9 \times \ldots \times 4\times 3 = 11! / 2$ ways; the result is $$ p=\frac{12\cdot 220\cdot 11!}{2\cdot 12^{12}}=\frac{105875}{17915904}\approx 0.00590955, $$ or about $0.6\%$ of the time.