Equation of tangent PT of the circle $$x^2+y^2=4$$ at $$\mathrm{P}(\sqrt{3}, 1)$$ is

$$\Rightarrow \quad \sqrt{3} x+y=4$$

Given, L is a line perpendicular to PT

$$\therefore \quad \mathrm{L} \equiv x-\sqrt{3} y=\lambda$$

Also given L is the tangent of circle

$$(x-3)^2+y^2=1$$

$$\begin{array}{ll} \Rightarrow & 1=\frac{|3-\sqrt{3} \cdot 0-\lambda|}{\sqrt{1^2+(-\sqrt{3})^2}} \\ \Rightarrow & |3-\lambda|=2 \\ \Rightarrow & 3-\lambda= \pm 2 \\ \Rightarrow & \lambda=1,5 \end{array}$$

Hence, the possible equation of line L are $$x-\sqrt{3} y=1$$ and $$x-\sqrt{3} y=5$$.