If $$\sin \theta = 3x  and  \cos \theta = \frac{3}{x}$$, (x ≠ 0) then the value of $$6\left(x^2 + \frac{1}{x^2}\right)$$ is:

Given, $$\sin\theta=3x$$ and $$\cos\theta=\frac{\ 3}{x}$$

We know that, $$\sin^2\theta+\cos^2\theta=1$$

$$=$$> $$\left(3x\right)^2+\left(\frac{\ 3}{x}\right)^2=1$$

$$=$$> $$9x^2+\frac{\ 9}{x^2}=1$$

$$=$$> $$9\left(x^2+\frac{\ 1}{x^2}\right)=1$$

$$=$$> $$x^2+\frac{\ 1}{x^2}=\frac{1}{9}$$

$$=$$> $$6\left(x^2+\frac{\ 1}{x^2}\right)=\frac{6}{9}$$

$$=$$> $$6\left(x^2+\frac{\ 1}{x^2}\right)=\frac{2}{3}$$

Hence, the correct answer is Option B