If $$\left[\left\{\left(\frac{2}{3}\right)^3\right\}^{(2x + 3)}\right]^{\frac{-3}{4}} = \left[\left\{\left(\frac{2}{3}\right)^{\frac{2}{3}}\right\}^{(3x + 7)}\right]^{\frac{-6}{5}}$$ then the value of $$\sqrt{2-42x}$$ is:

Given,

$$\left[\left\{\left(\frac{2}{3}\right)^3\right\}^{(2x + 3)}\right]^{\frac{-3}{4}} = \left[\left\{\left(\frac{2}{3}\right)^{\frac{2}{3}}\right\}^{(3x + 7)}\right]^{\frac{-6}{5}}$$

$$=$$>  $$\left\{\left(\frac{2}{3}\right)^3\right\}^{\frac{-3}{4}(2x+3)}=\left\{\left(\frac{2}{3}\right)^{\frac{2}{3}}\right\}^{\frac{-6}{5}(3x+7)}$$

$$=$$>  $$\left(\frac{2}{3}\right)^{\frac{-9}{4}(2x+3)}=\left(\frac{2}{3}\right)^{\frac{-12}{15}(3x+7)}$$

$$=$$>  $$\frac{-9}{4}(2x+3)=\frac{-12}{15}(3x+7)$$

$$=$$>  $$45(2x+3)=16(3x+7)$$

$$=$$>  $$90x+135=48x+112$$

$$=$$>  $$42x=-23$$

$$\therefore\ \sqrt{2-42x}=\sqrt{2-\left(-23\right)}=\sqrt{2+23}=\sqrt{25}=5$$

Hence, the correct answer is Option B