If $$\cos x = \frac{-\sqrt3}{2}  and  \pi < x < \frac{3\pi}{2}$$, then the value of $$2 \cot^2 x + 3 \cosec^2 x$$ is:

Given,  $$\cos x = \frac{-\sqrt3}{2}$$

$$=$$>  $$\sec x=\frac{-2}{\sqrt{3}}$$

$$\therefore\ $$ $$2\cot^2x+3\operatorname{cosec}^2x=\frac{2}{\tan^2x}+\frac{3}{\sin^2x}$$

$$=\frac{2}{\sec^2x-1}+\frac{3}{1-\cos^2x}$$

$$=\frac{2}{\left(\frac{-2}{\sqrt{3}}\right)^2-1}+\frac{3}{1-\left(\frac{\sqrt{3}}{2}\right)^2}$$

$$=\frac{2}{\frac{4}{3}-1}+\frac{3}{1-\frac{3}{4}}$$

$$=\frac{2}{\frac{1}{3}}+\frac{3}{\frac{1}{4}}$$

$$=6+12$$

$$=18$$

Hence, the correct answer is Option C