The problem can be done by straightforward calculation.
Initially no one is shooting at C, so at some point A or B will be hit. At that point either A and C survive, B and C survive, or C alone survives (if A and B are hit simultaneously).
A and C survive. This occurs when A hits B before B or C hits A. The probability that all three miss on any given turn is $\frac12\cdot\frac23\cdot\frac56=\frac5{18}$, and the probability that A hits and the other two miss on any given turn is also $\frac12\cdot\frac23\cdot\frac56=\frac5{18}$, so this event occurs on turn $n$ with probability $\left(\frac5{18}\right)^n$. The probability that it occurs at all is therefore $$\sum_{n\ge 1}\left(\frac5{18}\right)^n=\frac{\frac5{18}}{1-\frac5{18}}=\frac5{13}\;.$$
B and C survive. This occurs when B or C hits A before A hits B. As before, the probability that all three miss on any given turn is $\frac12\cdot\frac23\cdot\frac56=\frac5{18}$. The probability that neither B nor C hits on a given turn is $\frac23\cdot\frac56=\frac59$, so the probability that A misses and at least one of B and C hits is $\frac12\left(1-\frac59\right)=\frac29$. This event therefore occurs on turn $n$ with probability $$\left(\frac5{18}\right)^{n-1}\left(\frac29\right)\;,$$ and the probability that it occurs at all is $$\frac29\sum_{n\ge 1}\left(\frac5{18}\right)^{n-1}=\frac29\sum_{n\ge 0}\left(\frac5{18}\right)^n=\frac{\frac29}{1-\frac5{18}}=\frac4{13}\;.$$
C alone survives. This occurs when A hits B and, on the same turn, B or C hits A. Since the probability that A hits on the $n$ turn is the same as the probability that A misses, the calculation for this case is identical to that for the previous case, and this case therefore occurs with probability $\frac4{13}$. (Alternatively, of course, one can simply note that the probability must be $1-\frac5{13}-\frac4{13}=\frac4{13}$.)
Now consider the outcome of case (1). At this point A and C are shooting at each other. The probability that both miss on any given turn is $\frac12\cdot\frac56=\frac5{12}$, and the probability that A hits and C misses on any given turn is the same, so A becomes the sole survivor on turn $n$ with probability $\left(\frac5{12}\right)^n$. Should the battle reach this case, then, the probability that A is the sole survivor is
$$\sum_{n\ge 1}\left(\frac5{12}\right)^n=\frac{\frac5{12}}{1-\frac5{12}}=\frac57\;.$$
The probability that A misses and C hits on any given turn is $\frac12\cdot\frac16=\frac1{12}$, so the probability that C becomes the sole survivor on turn $n$ is $\left(\frac5{12}\right)^{n-1}\left(\frac1{12}\right)$, and the overall probability that C becomes the sole survivor, given that the battle reaches this case, is
$$\frac1{12}\sum_{n\ge 1}\left(\frac5{12}\right)^{n-1}=\frac1{12}\sum_{n\ge 0}\left(\frac5{12}\right)^n=\frac{\frac1{12}}{1-\frac5{12}}=\frac17\;.$$
Finally, once this case is reached there is a probability of $1-\frac57-\frac17=\frac17$ that A and C will kill each other, leaving no survivors.
The analysis for case (2) is entirely similar. The probability that B is the sole survivor is
$$\left(\frac13\cdot\frac56\right)\sum_{n\ge 1}\left(\frac23\cdot\frac56\right)^{n-1}=\frac{\frac5{18}}{1-\frac59}=\frac58\;,$$
the probability that C is the sole survivor is
$$\left(\frac23\cdot\frac16\right)\sum_{n\ge 1}\left(\frac23\cdot\frac56\right)^{n-1}=\frac{\frac19}{1-\frac59}=\frac14\;,$$
and the probability that there are no survivors is $1-\frac58-\frac14=\frac18$.
The overall probabilities of survival are therefore
$$\begin{align*}
&\text{A:}\quad\frac5{13}\cdot\frac57=\frac{25}{91}\\\\
&\text{B:}\quad\frac4{13}\cdot\frac58=\frac5{26}\\\\
&\text{C:}\quad\frac5{13}\cdot\frac17+\frac4{13}\cdot\frac14+\frac4{13}=\frac{40}{91}\;.
\end{align*}$$
As a quick check, the missing $\frac{17}{182}$ should be the probability that none of them survives; since this is
$$\frac5{13}\cdot\frac17+\frac4{13}\cdot\frac18\;,$$
which is indeed $\frac{17}{182}$.
