Let A, B and C be three events, which are pair-wise independent and $$\bar{E}$$ denotes the complement of an event E. If $$P(A \cap B \cap C) = 0$$ and $$P(C) > 0$$, then $$P\left[(\bar{A} \cap \bar{B}) | C\right]$$ is equal to:

We have three events $$A,\;B,\;C$$ that are pair-wise independent, so by definition we know

$$P(A\cap B)=P(A)\,P(B),\qquad P(A\cap C)=P(A)\,P(C),\qquad P(B\cap C)=P(B)\,P(C).$$

It is given that the simultaneous occurrence of all three events is impossible, that is

$$P(A\cap B\cap C)=0,$$

and also that $$P(C)>0.$$

We are asked to find the conditional probability

$$P\!\left[(\bar A\cap\bar B)\mid C\right].$$

By the definition of conditional probability,

$$P\!\left[(\bar A\cap\bar B)\mid C\right]= \frac{P\!\big((\bar A\cap\bar B)\cap C\big)}{P(C)}.$$

We now compute the numerator $$P((\bar A\cap\bar B)\cap C).$$ Observe that

$$(\bar A\cap\bar B)\cap C = C\setminus\big((A\cup B)\cap C\big) = C-\big((A\cap C)\cup(B\cap C)\big).$$

Using the addition rule for probabilities, we have

$$P\!\big((A\cap C)\cup(B\cap C)\big) = P(A\cap C)+P(B\cap C)-P(A\cap B\cap C).$$

Substituting the given information and the results from pair-wise independence:

$$P(A\cap C)=P(A)P(C),\qquad P(B\cap C)=P(B)P(C),\qquad P(A\cap B\cap C)=0.$$

So

$$P\!\big((A\cap C)\cup(B\cap C)\big) = P(A)P(C)+P(B)P(C)-0 = P(C)\big(P(A)+P(B)\big).$$

Hence

$$P\big((\bar A\cap\bar B)\cap C\big) = P(C)-P\!\big((A\cap C)\cup(B\cap C)\big) = P(C)-P(C)\big(P(A)+P(B)\big).$$

Taking $$P(C)$$ common, we get

$$P\big((\bar A\cap\bar B)\cap C\big) = P(C)\big[1-P(A)-P(B)\big] = P(C)\big[(1-P(A))-P(B)\big].$$

Recalling that $$1-P(A)=P(\bar A),$$ we have

$$P\big((\bar A\cap\bar B)\cap C\big) = P(C)\big[P(\bar A)-P(B)\big].$$

Now divide by $$P(C)$$ to obtain the desired conditional probability:

$$P\!\left[(\bar A\cap\bar B)\mid C\right] = \frac{P(C)\big[P(\bar A)-P(B)\big]}{P(C)} = P(\bar A)-P(B).$$

Thus, the required value equals $$P(\bar A)-P(B),$$ which matches Option A.

Hence, the correct answer is Option A.