Let $$S = \{x \in R : x \geq 0$$ & $$2|\sqrt{x} - 3| + \sqrt{x}(\sqrt{x} - 6) + 6 = 0\}$$. Then S:

We begin by observing that the unknown appears inside a square-root, so it is convenient to set $$t=\sqrt{x}\,.$$

Because $$x\ge 0,$$ we automatically have $$t\ge 0.$$ With this substitution the given condition

$$2\lvert\sqrt{x}-3\rvert+\sqrt{x}\,(\sqrt{x}-6)+6=0$$

becomes a purely quadratic-absolute value equation in $$t$$:

$$2\lvert t-3\rvert+t(t-6)+6=0.$$

Simplifying the second term first, we write

$$t(t-6)=t^{2}-6t.$$

Hence the complete equation is

$$2\lvert t-3\rvert+t^{2}-6t+6=0. \quad -(1)$$

The absolute value necessitates splitting into two cases.

Case 1: $$t\ge 3$$. Here $$\lvert t-3\rvert=t-3.$$ Substituting into (1) we get

$$2(t-3)+t^{2}-6t+6=0.$$

Expanding the brackets,

$$2t-6+t^{2}-6t+6=0.$$

Combining like terms,

$$t^{2}-4t=0.$$

Factoring,

$$t(t-4)=0.$$

This quadratic yields two possible roots:

$$t=0\quad\text{or}\quad t=4.$$

However, in the present case we imposed $$t\ge 3,$$ so only $$t=4$$ survives. Translating back,

$$x=t^{2}=4^{2}=16.$$

Case 2: $$0\le t<3$$. Here $$\lvert t-3\rvert=3-t.$$ Substituting into (1) we have

$$2(3-t)+t^{2}-6t+6=0.$$

Expanding,

$$6-2t+t^{2}-6t+6=0.$$

Combining constants and like terms,

$$t^{2}-8t+12=0.$$

To solve the quadratic, we use the standard formula $$t=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$ with $$a=1,\;b=-8,\;c=12.$$ This gives

$$t=\dfrac{8\pm\sqrt{64-48}}{2}=\dfrac{8\pm\sqrt{16}}{2}=\dfrac{8\pm4}{2}.$$ So,

$$t=6\quad\text{or}\quad t=2.$$

But the current case demands $$t<3,$$ hence only $$t=2$$ is admissible. Converting back to $$x$$ gives

$$x=t^{2}=2^{2}=4.$$

Combining both cases, the set $$S$$ is

$$S=\{4,\;16\}.$$

This set clearly has exactly two distinct non-negative elements.

Hence, the correct answer is Option D.