For a differentiable function $$f : \mathbb{R} \to \mathbb{R}$$, suppose $$f'(x) = 3f(x) + \alpha$$, where $$\alpha \in \mathbb{R}$$, $$f(0) = 1$$ and $$\lim_{x \to -\infty} f(x) = 7$$. Then $$9f(-\log_e 3)$$ is equal to ________

$$f'(x)=3f(x)+\alpha, \alpha\in\mathbb{R}$$ with the conditions $$f(0)=1$$ and $$\displaystyle\lim_{x\to-\infty}f(x)=7$$.
$$f'(x)-3f(x)=\alpha.$$

The integrating factor (I.F.) for $$y'-3y=0$$ is $$e^{-3x}$$, because
$$\text{I.F.}=e^{\int -3\,dx}=e^{-3x}.$$
$$e^{-3x}f'(x)-3e^{-3x}f(x)=\alpha e^{-3x}.$$

The left side is the derivative of the product $$f(x)e^{-3x}$$, so
$$\frac{d}{dx}\bigl[f(x)e^{-3x}\bigr]=\alpha e^{-3x}.$$

Integrate with respect to $$x$$:
$$f(x)e^{-3x}=\int \alpha e^{-3x}\,dx + C$$
$$\Rightarrow f(x)e^{-3x}=\alpha\left(-\frac{1}{3}\right)e^{-3x}+C.$$

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

Use the initial condition $$f(0)=1$$:
$$1=-\frac{\alpha}{3}+C e^{0}\quad\Longrightarrow\quad C=1+\frac{\alpha}{3}\, . \qquad -(1)$$

Use the limit condition $$\displaystyle\lim_{x\to-\infty}f(x)=7$$.

Since $$e^{3x}\to 0$$ as $$x\to-\infty$$, the exponential term vanishes, leaving
$$7=-\frac{\alpha}{3}\quad\Longrightarrow\quad \alpha=-21.$$

Substitute $$\alpha=-21$$ into $$(1)$$ to find $$C$$:
$$C=1+\frac{-21}{3}=1-7=-6.$$
$$f(x)=-6e^{3x}+7.$$

Evaluate at $$x=-\log_e 3$$ (note: $$-\log_e 3=\ln\!\bigl(\tfrac{1}{3}\bigr)$$):
$$e^{3x}=e^{3(-\log_e 3)}=e^{-\log_e 3^{3}}=3^{-3}=\frac{1}{27}.$$
$$f(-\log_e 3)=-6\left(\frac{1}{27}\right)+7=-\frac{6}{27}+7=-\frac{2}{9}+7=\frac{-2+63}{9}=\frac{61}{9}.$$
$$9f(-\log_e 3)=9\left(\frac{61}{9}\right)=61.$$

Thus the required value is $$\mathbf{61}$$.