The number of stereoisomers formed in a reaction of $$[\pm] PhC(=O)C(OH)(CN)Ph$$ with HCN is

We are given the racemic compound $$[\pm]\; \text{PhC}(=\text{O})\text{C}(\text{OH})(\text{CN})\text{Ph}$$, which is a cyanohydrin with structure $$\text{Ph}-\overset{\displaystyle O}{\overset{\displaystyle \|}{C}}-\overset{\displaystyle OH \quad CN}{\overset{\displaystyle | \quad\;\; |}{C}}-\text{Ph}$$. We have a carbonyl group ($$\text{C}=\text{O}$$) on one carbon and a chiral centre on the adjacent carbon that bears $$-\text{OH}$$, $$-\text{CN}$$, $$-\text{Ph}$$, and the bond to the carbonyl carbon. The $$[\pm]$$ indicates both enantiomers (R and S) of this chiral centre are present.

When HCN adds to the $$\text{C}=\text{O}$$ group, the cyanide nucleophile attacks the carbonyl carbon. This converts the $$\text{C}=\text{O}$$ into a $$\text{C}(\text{OH})(\text{CN})$$ group, creating a new chiral centre. The product is therefore $$\text{Ph}-\text{C}(\text{OH})(\text{CN})-\text{C}(\text{OH})(\text{CN})-\text{Ph}$$, which has two chiral centres.

Now we observe something important: both chiral carbons carry the same set of four groups — $$\text{Ph}$$, $$\text{OH}$$, $$\text{CN}$$, and a bond to the other chiral carbon (which itself bears $$\text{Ph}$$, $$\text{OH}$$, $$\text{CN}$$). Because both stereocentres are substituted identically, the molecule can potentially possess an internal plane of symmetry when the two centres have opposite configurations.

We enumerate the stereoisomers. With two stereocentres of identical substitution, the possibilities are: the $$(R,R)$$ form, the $$(S,S)$$ form (which is the mirror image of $$(R,R)$$), and the $$(R,S)$$ form. The $$(R,S)$$ configuration is the same as $$(S,R)$$ because the molecule has an internal mirror plane — this is the meso compound. So we get exactly three stereoisomers: one pair of enantiomers $$(R,R)$$ and $$(S,S)$$, plus the meso form $$(R,S)$$.

Hence, the correct answer is $$\boxed{3}$$.