We need the values of $a\in \mathbb R\setminus\{1\}$ for which both roots of [ (1-a)x^2+2(a-3)x+9=0 ] are positive.

Let the roots be $r_1,r_2$.

1. Conditions for both roots to be positive

For a quadratic $Ax^2+Bx+C=0$, both roots are positive iff:

1. Roots are real: $D\ge 0$
2. Sum of roots is positive: $r_1+r_2>0$
3. Product of roots is positive: $r_1r_2>0$

Here, [ A=1-a,\quad B=2(a-3),\quad C=9. ]

By Vieta, [ r_1+r_2=-\frac{B}{A}=-\frac{2(a-3)}{1-a}=\frac{2(a-3)}{a-1}, ] [ r_1r_2=\frac{C}{A}=\frac{9}{1-a}. ]

2. Product condition

For both roots positive, [ r_1r_2>0 \implies \frac{9}{1-a}>0. ] Since $9>0$, this gives [ 1-a>0 \implies a<1. ]

3. Sum condition

We need [ r_1+r_2=\frac{2(a-3)}{a-1}>0. ] For $a<1$, denominator $a-1<0$. Also $a-3<0$ for all $a<1$. So numerator and denominator are both negative, hence the fraction is positive. Thus for every $a<1$, the sum condition is automatically satisfied.

4. Discriminant condition

Now require real roots: [ D=B^2-4AC=[2(a-3)]^2-4(1-a)(9). ] Compute: [ D=4(a-3)^2-36(1-a). ] Expand: [ (a-3)^2=a^2-6a+9, ] so [ D=4(a^2-6a+9)-36+36a ] [ =4a^2-24a+36-36+36a ] [ =4a^2+12a=4a(a+3). ] Thus, [ D\ge 0 \iff 4a(a+3)\ge 0 \iff a(a+3)\ge 0. ] So, [ a\le -3 \quad \text{or} \quad a\ge 0. ]

5. Combine conditions

We need both:

• $a<1$
• $a\le -3$ or $a\ge 0$

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

This matches the form [ (-\infty,-\alpha]\cup[\beta,\gamma). ] So, [ \alpha=3,\quad \beta=0,\quad \gamma=1. ]

Therefore, [ 2\alpha+\beta+\gamma=2(3)+0+1=7. ]

6. Comparison with stored answer

Derived answer = $7$. Stored correct answer = $7$. They agree.