If the set of all $a \in \mathbb{R} \setminus \{1\}$, 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:

Step 1: Analyze conditions for positive roots

For a quadratic equation $Ax^2 + Bx + C = 0$ to have positive roots, three conditions must be met: the discriminant $D \ge 0$, the sum of the roots $-B/A > 0$, and the product of the roots $C/A > 0$. In our given equation, $A = (1-a)$, $B = 2(a-3)$, and $C = 9$.

Step 2: Apply Discriminant Condition

First, we apply the discriminant condition $D \ge 0$. Substituting the coefficients, we get $4(a-3)^2 - 36(1-a) \ge 0$. Dividing by 4 and expanding, we have $(a^2 - 6a + 9) - 9 + 9a \ge 0$, which simplifies to $a^2 + 3a \ge 0$. Factoring gives $a(a+3) \ge 0$. This inequality holds when $a \in (-\infty, -3] \cup [0, \infty)$.

Step 3: Apply Sum of Roots Condition

Next, we apply the sum of roots condition, $-B/A > 0$. Substituting the coefficients, we get $-2(a-3)/(1-a) > 0$. This inequality can be rewritten as $2(a-3)/(a-1) > 0$. This inequality holds when $a \in (-\infty, 1) \cup (3, \infty)$.

Step 4: Apply Product of Roots Condition

Finally, we apply the product of roots condition, $C/A > 0$. Substituting the coefficients, we get $9/(1-a) > 0$. This inequality holds when $1-a > 0$, which means $a < 1$. So, $a \in (-\infty, 1)$.

Step 5: Find the intersection of all conditions

To find the set of all $a$ for which the roots are positive, we need to find the intersection of the solution sets from all three conditions. The intersection is $(-\infty, -3]$. Comparing this with the given form $(-\infty, -\alpha] \cup [\beta, \gamma)$, we find that $-\alpha = -3$, so $\alpha = 3$. The union part $[\beta, \gamma)$ is empty, which means $\beta$ and $\gamma$ are not defined in this context, or we can consider the interval to be $(-\infty, -3]$ and the union part to be empty. However, the problem implies a specific form. Let's re-evaluate the intersection carefully. The intersection of $(-\infty, -3] \cup [0, \infty)$ and $(-\infty, 1) \cup (3, \infty)$ is $(-\infty, -3] \cup [0, 1) \cup (3, \infty)$. Intersecting this with $(-\infty, 1)$ gives $(-\infty, -3] \cup [0, 1)$.

Step 6: Determine $\alpha, \beta, \gamma$ and calculate $2\alpha + \beta + \gamma$

From the intersection $(-\infty, -3] \cup [0, 1)$, we compare it with the given form $(-\infty, -\alpha] \cup [\beta, \gamma)$. This implies $-\alpha = -3$, so $\alpha = 3$. Also, $\beta = 0$ and $\gamma = 1$. Now we can calculate $2\alpha + \beta + \gamma = 2(3) + 0 + 1 = 6 + 0 + 1 = 7$.

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