Given,

$$2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$$

Case 1 :

When $$\sqrt x - 3 \ge 0,$$ then equation becomes

$$2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$$

$$ \Rightarrow \,\,\,\,2\sqrt x - 6 + x - 6\sqrt x + 6 = 0$$

$$ \Rightarrow \,\,\,\,x\, - 4\sqrt x = 0$$

$$ \Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 4} \right) = 0$$

$$\therefore\,\,\,$$ $$\sqrt x = 0,4$$

but as $$\sqrt x - 3 \ge 0$$ or $$\sqrt x \ge 3$$ then $$\sqrt x \ne 0$$

$$\therefore\,\,\,$$ $$\sqrt x = 4$$ value is possible.

Case 2 :

When $$\sqrt x - 3 < 0$$ or $$\sqrt x < 3.$$ There equation becomes

$$ - 2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$$

$$ \Rightarrow \,\,\,\,\, - 2\sqrt x + 6 + x - 6\sqrt x + 6 = 0$$

$$ \Rightarrow \,\,\,\,x - 8\sqrt x + 12 = 0$$

$$ \Rightarrow \,\,\,\,x - 6\sqrt x - 2\sqrt x + 12 = 0$$

$$ \Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 6} \right) - 2\left( {\sqrt x - 6} \right) = 0$$

$$ \Rightarrow \,\,\,\,\left( {\sqrt x - 2} \right)\left( {\sqrt x - 6} \right) = 0$$

$$\therefore\,\,\,$$ $$\sqrt x = 2,6$$

as $$\sqrt x < 3$$ so $$\sqrt x \ne 6$$

$$\therefore\,\,\,$$ $$\sqrt x = 2$$ is possible.

So, total possible value of $$\sqrt x = 2,4$$

or for x possible values are 4, 16.

$$\therefore\,\,\,$$ Set S contains exactly two elements 4 and 16.