Benzaldehyde on reaction with acetophenone in the presence of sodium hydroxide solution gives

Reactant $A$ of the above reaction is
$(A)$ + $\begin{matrix} CH_2-OH \\ | \\ CH_2-OH \end{matrix}$ $\xrightarrow{\text{Pyridine}} \begin{matrix} CH_2-O \\ | \\ CH_2-O \end{matrix} > C = O$

Consider the given reaction,the product $'X'$ is:

When $(X)$ reacts with $(Y)$ in the presence of dilute $(Z)$ solution,$3-$hydroxybutanal is formed. What are $(X), (Y),$ and $(Z)$?

$PhMgBr$ [Excess] + $CH_3COCl \xrightarrow{H_3O^+}$
The final product of the above reaction is:

In the following compounds,the ones that give positive iodoform test are:
$(I)$ $CH_3CH_2COCH_2CH_3$
$(II)$ $CH_3CH(OH)CH_3$
$(III)$ $1$-indanone
$(IV)$ $CH_3CH_2COCH_3$
$(V)$ $PhCOPh$
$(VI)$ $PhCOCH_3$