$$\begin{aligned} & f^{\prime}(x)=3 f(x)+\alpha \\ & \Rightarrow \frac{d y}{3 y+\alpha}=d x \\ & \Rightarrow \frac{1}{3} \ln (3 y+\alpha)=x+C \\ & y(0)=1 \Rightarrow C=\frac{1}{3} \ln (3+\alpha) \\ & \frac{1}{3} \ln \left(\frac{3 y+\alpha}{3+\alpha}\right)=x \\ & \Rightarrow y=\frac{1}{3}\left((3+\alpha) e^{3 x}-\alpha\right)=f(x) \\ & \lim _{x \rightarrow-\infty} f(x)=7 \Rightarrow \alpha=-21 \\ & \Rightarrow f(x)=7-6 e^{3 x} \\ & 9 f(-\ln 3)=61 \end{aligned}$$