1. Given differential equation

   $f'(x)=3f(x)+\alpha$

   with

   $f(0)=1, \qquad \lim_{x\to -\infty} f(x)=7.$

2. Solve the linear differential equation

   Rewrite it as

   $f'(x)-3f(x)=\alpha.$

   A constant particular solution can be taken as $f(x)=k$. Then

   $0-3k=\alpha \implies k=-\frac{\alpha}{3}.$

   So the general solution is

   $f(x)=Ce^{3x}-\frac{\alpha}{3}.$

3. Use the limit condition

   As $x\to -\infty$,

   $e^{3x}\to 0.$

   Hence

   $\lim_{x\to -\infty} f(x)= -\frac{\alpha}{3}.$

   But this limit is given as $7$, so

   $-\frac{\alpha}{3}=7 \implies \alpha=-21.$

   Therefore,

   $f(x)=Ce^{3x}+7.$

4. Use the initial condition

   Since $f(0)=1$,

   $C+7=1 \implies C=-6.$

   Thus

   $f(x)=7-6e^{3x}.$

5. Evaluate at $x=-\ln 3$

   $e^{3(-\ln 3)}=e^{-3\ln 3}=3^{-3}=\frac{1}{27}.$

   So

   $f(-\ln 3)=7-6\left(\frac{1}{27}\right)=7-\frac{2}{9}=\frac{61}{9}.$

6. Compute the required value

   $9f(-\ln 3)=9\cdot \frac{61}{9}=61.$

Therefore, the required integer is

$\boxed{61}.$