If $$x = 5 - \sqrt {21} {\text{,}}$$   then the value of $$\frac{{\sqrt x }}{{\sqrt {32 - 2x} - \sqrt {21} }}$$    is?

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