Out of all the patients in a hospital 89% are found to be suffering from heart ailment and 98% are suffering from lungs infection. If $$K$$% of them are suffering from both ailments, then $$K$$ can not belong to the set:

Let us denote by $$H$$ the set of patients having a heart ailment and by $$L$$ the set of patients having a lung infection.

We are told that $$P(H)=89\%$$ and $$P(L)=98\%$$, where $$P(\cdot)$$ represents the percentage of the total number of patients.

For two sets, the Inclusion-Exclusion Principle states first:

$$P(H\cup L)=P(H)+P(L)-P(H\cap L).$$

Secondly, because a percentage cannot exceed the whole, we have the bound:

$$P(H\cup L)\le 100\%.$$

Combining these two facts, we write

$$P(H)+P(L)-P(H\cap L)\le 100.$$

Substituting the given values,

$$89+98-K\le 100,$$

where $$K$$ is exactly $$P(H\cap L)$$, the percentage suffering from both ailments.

Now we simplify step by step:

$$187-K\le 100,$$

so

$$-K\le 100-187,$$

hence

$$-K\le -87,$$

and multiplying both sides by $$-1$$ (remembering to reverse the inequality sign) we obtain

$$K\ge 87.$$

Next, we recall another elementary bound: the intersection of two sets can never exceed either of the individual sets. Symbolically,

$$P(H\cap L)\le \min\{P(H),P(L)\}.$$

Because $$P(H)=89\%$$ and $$P(L)=98\%,$$ the smaller of the two is $$89\%,$$ so

$$K\le 89.$$

Putting the two inequalities together, we get the precise interval

$$87\le K\le 89.$$

Thus the only possible integral values of $$K$$ are $$87, 88, 89.$$ Any value outside this closed interval is impossible.

Now we inspect each given option:

Option A: $$\{79,81,83,85\}$$ - none of these values lie in $$[87,89].$$

Option B: $$\{84,87,90,93\}$$ - the element $$87$$ does lie in $$[87,89].$$

Option C: $$\{80,83,86,89\}$$ - the element $$89$$ does lie in $$[87,89].$$

Option D: $$\{84,86,88,90\}$$ - the element $$88$$ does lie in $$[87,89].$$

Therefore, the only set in which $$K$$ cannot possibly fall is Option A.

Hence, the correct answer is Option A.