The conversion of hydroxyapatite occurs due to presence of F$$^-$$ ions in water. The correct formula of hydroxyapatite is:

We begin by recalling a very well-known fact from elementary bio-inorganic chemistry: the main inorganic constituent of tooth enamel is hydroxyapatite. Its accepted chemical formula is written as

$$Ca_{10}(PO_4)_6(OH)_2.$$

Because questions in JEE often rewrite such formulas in a factored or “mixed salt” form, we now try to re-express $$Ca_{10}(PO_4)_6(OH)_2$$ as a product of two simpler familiar units so that it will resemble one of the options given.

First, we notice that the ionic compound $$Ca_3(PO_4)_2$$, known as tricalcium phosphate, is present inside the overall formula. To see how many such units are hidden in hydroxyapatite, we compare the numbers of each atom.

We write:

Let $$x$$ be the number of $$Ca_3(PO_4)_2$$ units and let $$y$$ be the number of $$Ca(OH)_2$$ units that together produce $$Ca_{10}(PO_4)_6(OH)_2$$.

So we want

$$x\bigl[Ca_3(PO_4)_2\bigr] + y\bigl[Ca(OH)_2\bigr] \;=\; Ca_{10}(PO_4)_6(OH)_2.$$

We now equate the quantity of each type of atom on both sides.

Calcium atoms:

$$3x + y = 10 \quad\text{(1)}$$

Phosphate groups:

Because each $$Ca_3(PO_4)_2$$ carries 2 phosphate groups $$PO_4^{3-}$$, the total number of phosphate groups equals $$2x$$, and we require

$$2x = 6 \quad\text{(2)}$$

Solving equation (2) first, we divide both sides by 2 and obtain

$$x = 3.$$

Substituting $$x = 3$$ into equation (1) gives

$$3(3) + y = 10 \;\;\Longrightarrow\;\; 9 + y = 10 \;\;\Longrightarrow\;\; y = 1.$$

Hence we have exactly three units of $$Ca_3(PO_4)_2$$ and one unit of $$Ca(OH)_2$$ in a single formula unit of hydroxyapatite. Writing that product out explicitly, we get

$$\bigl[3Ca_3(PO_4)_2 \cdot Ca(OH)_2\bigr],$$

which is an alternative but completely equivalent way of expressing $$Ca_{10}(PO_4)_6(OH)_2.$$

We compare this result with the options provided:

A. $$[Ca_3(PO_4)_2 \cdot CaF_2]$$ — contains fluoride, not hydroxide; incorrect.

B. $$[3Ca_3(PO_4)_2 \cdot Ca(OH)_2]$$ — exactly matches the expression we derived; correct.

C. $$[3Ca_3(PO_4)_2 \cdot CaF_2]$$ — hydroxide replaced by fluoride; incorrect.

D. $$[3Ca(OH)_2 \cdot CaF_2]$$ — phosphate totally missing; incorrect.

So, the conversion of hydroxyapatite in the presence of $$F^-$$ ions indeed starts from $$[3Ca_3(PO_4)_2 \cdot Ca(OH)_2]$$.

Hence, the correct answer is Option B.