Let $$E^C$$ denote the complement of an event $$E$$. Let $$E_1$$, $$E_2$$ and $$E_3$$ be any pairwise independent events with $$P(E_1) > 0$$ and $$P(E_1 \cap E_2 \cap E_3) = 0$$ then $$P\left((E_2^C \cap E_3^C)/E_1\right)$$ is equal to:

We have to determine the conditional probability $$P\!\left((E_2^{\,C}\cap E_3^{\,C})\,\big/\,E_1\right)\!.$$

First recall the definition of conditional probability:

$$P(A/B)=\dfrac{P(A\cap B)}{P(B)},\qquad\text{provided }P(B)>0.$$

Taking $$A=(E_2^{\,C}\cap E_3^{\,C})\quad\text{and}\quad B=E_1,$$ we obtain

$$P\!\left((E_2^{\,C}\cap E_3^{\,C})\big/ E_1\right)=\dfrac{P\!\left(E_1\cap E_2^{\,C}\cap E_3^{\,C}\right)}{P(E_1)}.$$

Hence our immediate task is to evaluate $$P\!\left(E_1\cap E_2^{\,C}\cap E_3^{\,C}\right).$$ Observe that

$$E_1\cap E_2^{\,C}\cap E_3^{\,C}=E_1\setminus(E_2\cup E_3),$$

so by the subtraction rule of probability,

$$P\!\left(E_1\cap E_2^{\,C}\cap E_3^{\,C}\right)=P(E_1)-P\!\left(E_1\cap(E_2\cup E_3)\right).$$

Using the formula for the probability of a union,

$$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y),$$

with $$X=E_1\cap E_2,\;Y=E_1\cap E_3,$$ we have

$$P\!\left(E_1\cap(E_2\cup E_3)\right)=P(E_1\cap E_2)+P(E_1\cap E_3)-P(E_1\cap E_2\cap E_3).$$

The problem states that $$P(E_1\cap E_2\cap E_3)=0,$$ so the last term vanishes. Therefore

$$P\!\left(E_1\cap(E_2\cup E_3)\right)=P(E_1\cap E_2)+P(E_1\cap E_3).$$

Because the events are pairwise independent, we can employ the independence property $$P(E_i\cap E_j)=P(E_i)\,P(E_j)$$ for every pair. Thus

$$P(E_1\cap E_2)=P(E_1)P(E_2),\qquad P(E_1\cap E_3)=P(E_1)P(E_3).$$

Substituting into the previous expression,

$$P\!\left(E_1\cap(E_2\cup E_3)\right)=P(E_1)P(E_2)+P(E_1)P(E_3).$$

Now place this back into the earlier subtraction expression:

$$P\!\left(E_1\cap E_2^{\,C}\cap E_3^{\,C}\right)=P(E_1)-\big[P(E_1)P(E_2)+P(E_1)P(E_3)\big].$$

Factoring out the common term $$P(E_1)$$ yields

$$P\!\left(E_1\cap E_2^{\,C}\cap E_3^{\,C}\right)=P(E_1)\Bigl[1-P(E_2)-P(E_3)\Bigr].$$

Returning to the conditional probability and dividing by $$P(E_1)\;(\text{which is}>0),$$ we get

$$P\!\left((E_2^{\,C}\cap E_3^{\,C})/E_1\right)=\dfrac{P(E_1)\bigl[1-P(E_2)-P(E_3)\bigr]}{P(E_1)}=1-P(E_2)-P(E_3).$$

Noting that the complement of $$E_3$$ has probability $$P(E_3^{\,C})=1-P(E_3),$$ we can rewrite the expression as

$$1-P(E_2)-P(E_3)=\bigl[1-P(E_3)\bigr]-P(E_2)=P(E_3^{\,C})-P(E_2).$$

This matches option D in the list provided.

Hence, the correct answer is Option D.