If the set of all $$a \in \mathbb{R} - \{1\}$$, for which the roots of the equation $$(1 - a)x^2 + 2(a - 3)x + 9 = 0$$ are positive is $$(-\infty, -\alpha] \cup [\beta, \gamma)$$, then $$2\alpha + \beta + \gamma$$ is equal to __________.

For the quadratic $$ (1-a)x^{2}+2(a-3)x+9=0 $$ let

$$A = 1-a, \; B = 2(a-3), \; C = 9.$$

For both roots to be real and positive we need three conditions:

1. Real roots: discriminant $$\Delta \ge 0$$.
2. Positive product: $$P = \dfrac{C}{A} \gt 0$$.
3. Positive sum: $$S = -\dfrac{B}{A} \gt 0.$$

Step 1 Discriminant
$$\Delta = B^{2}-4AC = \left[2(a-3)\right]^{2}-4(1-a)(9).$$

Simplify:

$$\Delta = 4(a-3)^{2}-36(1-a) = 4\big[(a-3)^{2}-9(1-a)\big]$$ $$= 4\big[a^{2}-6a+9-9+9a\big] = 4\big[a^{2}+3a\big] = 4a(a+3).$$

Thus $$\Delta \ge 0 \Longrightarrow a(a+3) \ge 0 \Longrightarrow a \le -3 \; \text{or} \; a \ge 0.$$

Step 2 Product of roots
$$P = \dfrac{C}{A} = \dfrac{9}{1-a}.$$

For $$P \gt 0$$ the denominator must be positive:

$$1-a \gt 0 \Longrightarrow a \lt 1.$$

Step 3 Sum of roots
$$S = -\dfrac{B}{A} = -\dfrac{2(a-3)}{1-a} = \dfrac{2(a-3)}{a-1}.$$

We need $$S \gt 0$$, so

$$\dfrac{a-3}{a-1} \gt 0.$$

The critical points are $$a = 1$$ and $$a = 3$$. Using a sign chart:

• For $$a \lt 1$$, both $$a-3$$ and $$a-1$$ are negative ⇒ ratio positive.
• For $$1 \lt a \lt 3$$, the ratio is negative.
• For $$a \gt 3$$, both factors are positive ⇒ ratio positive.

Hence $$S \gt 0$$ when $$a \lt 1$$ or $$a \gt 3$$ (note $$a \ne 1$$).

Step 4 Combine all conditions

• From product: $$a \lt 1.$$
• From sum: also satisfied for $$a \lt 1.$$
• From discriminant: $$a \le -3$$ or $$a \ge 0.$$
Intersecting with $$a \lt 1$$ gives

$$a \in (-\infty,-3] \;\; \cup \;\; [0,1).$$

Step 5 Identify $$\alpha,\beta,\gamma$$
The required set is given as $$(-\infty,-\alpha] \cup [\beta,\gamma).$$ Matching, we get $$\alpha = 3,\; \beta = 0,\; \gamma = 1.$

Step 6 Compute the expression
$$2$$\alpha + \beta + \gamma$$ = 2(3) + 0 + 1 = 7.$$

Therefore, the required value is $$\boxed{7}.$$