The probabilities of three events A, B and C are given $$P(A) = 0.6$$, $$P(B) = 0.4$$ and $$P(C) = 0.5$$. If $$P(A \cup B) = 0.8$$, $$P(A \cap C) = 0.3$$, $$P(A \cap B \cap C) = 0.2$$, $$P(B \cap C) = \beta$$ and $$P(A \cup B \cup C) = \alpha$$, where $$0.85 \leq \alpha \leq 0.95$$, then $$\beta$$ lies in the interval:

We have the individual probabilities

$$P(A)=0.6,\qquad P(B)=0.4,\qquad P(C)=0.5.$$

First, from the union of A and B we can determine their intersection. The addition (union) rule of two events is

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

Substituting the given numbers,

$$0.8 = 0.6 + 0.4 - P(A\cap B).$$

So

$$P(A\cap B)=0.6+0.4-0.8 = 0.2.$$

The triple intersection is given directly:

$$P(A\cap B\cap C)=0.2.$$

The intersection of A and C is also supplied:

$$P(A\cap C)=0.3.$$

Let us denote

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

To involve $$\beta$$ and $$\alpha$$ simultaneously, we invoke the general addition (union) rule for three events:

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

Substituting every known value step by step,

$$\alpha = 0.6 + 0.4 + 0.5 \;-\; 0.2 \;-\; 0.3 \;-\; \beta \;+\; 0.2.$$

Simplifying the numeric part carefully,

$$\alpha = 1.5 - 0.2 - 0.3 - \beta + 0.2 = 1.2 - \beta.$$

The problem states that $$\alpha$$ must satisfy

$$0.85 \le \alpha \le 0.95.$$

Replacing $$\alpha$$ by $$1.2-\beta$$, we obtain the compound inequality

$$0.85 \le 1.2 - \beta \le 0.95.$$

We solve each side separately.

Left-hand side:

$$0.85 \le 1.2 - \beta \;\;\Longrightarrow\;\; -\beta \le 1.2 - 0.85 = 0.35 \;\;\Longrightarrow\;\; \beta \le 0.35.$$

Right-hand side:

$$1.2 - \beta \le 0.95 \;\;\Longrightarrow\;\; -\beta \le 0.95 - 1.2 = -0.25 \;\;\Longrightarrow\;\; \beta \ge 0.25.$$

Combining the two inequalities we get

$$0.25 \le \beta \le 0.35.$$

Finally, we should check that this range is also permissible with elementary constraints such as

$$\beta \le P(B)=0.4,\qquad \beta \le P(C)=0.5,\qquad \beta \ge P(A\cap B\cap C)=0.2,$$

all of which are clearly satisfied by every value in the interval $$[0.25,0.35]$$.

Therefore $$\beta$$ must lie in the interval $$[0.25,0.35].$$

Among the given choices, this corresponds precisely to Option B.

Hence, the correct answer is Option B.