If $$2\cot\theta=3$$, then $$\frac{\sqrt{13}\cos\theta-3\tan\theta}{3\tan\theta+\sqrt{13}\sin\theta}$$ is:

Given, $$2\cot\theta=3$$

$$\Rightarrow$$  $$\cot\theta=\frac{3}{2}$$

$$\tan\theta=\frac{2}{3}$$

$$\operatorname{cosec}\theta\ =\sqrt{1+\cot^2\theta\ }=\sqrt{1+\left(\frac{3}{2}\right)2}=\sqrt{1+\frac{9}{4}}=\sqrt{\frac{13}{4}}=\frac{\sqrt{13}}{2}$$

$$\sin\theta\ =\frac{2}{\sqrt{13}}$$

$$\cos\theta\ =\sqrt{1-\sin^2\theta\ }=\sqrt{1-\frac{4}{13}}=\sqrt{1-\left(\frac{2}{\sqrt{13}}\right)^2}=\sqrt{\frac{9}{13}}=\frac{3}{\sqrt{13}}$$

$$\therefore\ $$ $$\frac{\sqrt{13}\cos\theta-3\tan\theta}{3\tan\theta+\sqrt{13}\sin\theta}=\frac{\sqrt{13}\left(\frac{3}{\sqrt{13}}\right)-3\left(\frac{2}{3}\right)}{3\left(\frac{2}{3}\right)+\sqrt{13}\left(\frac{2}{\sqrt{13}}\right)}$$

$$=\frac{3-2}{2+2}$$

$$=\frac{1}{4}$$

Hence, the correct answer is Option B