If x = 5 - 21 , then the value of x 32 - 2x - 21 is?

Answer: Option B

$$\begin{array}{rl}& x=5-\sqrt{21}\\ & 2x=10-2\sqrt{21}......(i)\\ & \Rightarrow 2x={\left(\sqrt{7}\right)}^2+{\left(\sqrt{3}\right)}^2-2\left(\sqrt{7}\right)\left(\sqrt{3}\right)\\ & \Rightarrow 2x={\left(\sqrt{7}-\sqrt{3}\right)}^2\\ & \Rightarrow x=\frac{1}{2}{\left(\sqrt{7}-\sqrt{3}\right)}^2\\ & \Rightarrow \sqrt{x}=\frac{1}{\sqrt{2}}\sqrt{{\left(\sqrt{7}-\sqrt{3}\right)}^2}\\ & \Rightarrow \sqrt{x}=\frac{1}{\sqrt{2}}\left(\sqrt{7}-\sqrt{3}\right)\\ & \therefore \frac{\sqrt{x}}{\sqrt{32-2x}-\sqrt{21}}\\ & =\frac{\sqrt{7}-\sqrt{3}}{\sqrt{2}\left[\sqrt{32-\left(10-2\sqrt{21}\right)}-\sqrt{21}\right]}\\ & =\frac{\sqrt{7}-\sqrt{3}}{\sqrt{2}\left[\sqrt{22+2\sqrt{21}}-\sqrt{21}\right]}\\ & =\frac{\sqrt{7}-\sqrt{3}}{\sqrt{2}\left[\sqrt{{\left(\sqrt{21}+1\right)}^2}-\sqrt{21}\right]}\\ & =\frac{\sqrt{7}-\sqrt{3}}{\sqrt{2}\left[\sqrt{21}+1-\sqrt{21}\right]}\\ & =\frac{\sqrt{7}-\sqrt{3}}{\sqrt{2}}\\ & =\frac{1}{\sqrt{2}}\left(\sqrt{7}-\sqrt{3}\right)\end{array}$$