I will assume you initially choose one of the three coins at random.
There are 6 equally likely possibilities for the side you might see: the "head side" of the HT coin, the tail side of the HT coin, one "head side" of the HH coin, the "other head side" of the HH coin, one tail side of the TT coin, and the other tail side of the TT coin.
Since you observed a tail, three of the six, equally likely, possibilities listed above are ruled out, and you know you either saw one side of the TT coin or you saw the tail side of the TH coin.
The probability that the other side of the coin is also a tail is 2/3.
Here's a different way to do it:
Let $A$ be the event the observed side is a tail, $X_{HH}$ be the event that you picked the HH coin, $X_{TT}$ be the event that you picked TT coin, and $X_{TH}$ be the event that you picked the TH coin.
The desired probability is $P(X_{TT}|A)$. We have:
$$\eqalign{
P(X_{TT}|A)&={P(A|X_{TT})P(X_{TT})\over P(A)}\cr
&={P(A|X_{TT})P(X_{TT})\over P(A|X_{HH})P(X_{HH}) + P(A|X_{HT})P(X_{HT})
+P(A|X_{TT})P(X_{TT})}\cr
&={ 1\cdot(1/3) \over 0\cdot(1/3)+(1/2)(1/3)+1\cdot(1/3) }\cr
&={1 \over (1/2)+1}\cr
&=2/3.
}
$$