Electrochemistry — Chemistry STD 12 Science — Question

$\underset{{{C}_{3}}{{H}_{6}}C{{l}_{2}}}{\mathop{A}}\,\,\xrightarrow[2.\,\,{{H}^{+}}\,\,(\min d)]{\begin{smallmatrix} 
 1\,.\,\,alcoholic\,\,KOH\,/\,\Delta  \\ 
 followed\,\,by\,\,NaN{{H}_{2}}\,/\,\Delta  
\end{smallmatrix}}$ $B\,\xrightarrow[{{H}_{2}}{{O}_{2}}/\overset{\Theta }{\mathop{O}}\,H]{{{B}_{2}}{{H}_{6}}}C$

$A\,\xrightarrow{aq.\,\,KOH\,(excess)}C$

($1$) For the synthesis of benzoic acid, the only $CORRECT$ combination is

$[A] (III) (iv) (R)$   $[B] (IV)(ii) (P)$   $[C] (I) (iv) (Q)$   $[D] (II) (i) (S)$

($2$)  The only CORRECT combination in which the reaction proceeds through radical mechanism is

$[A] (I)$ $(ii) (R)$     $[B] (II) (iii) (R)$     $[C] (III) (ii) (P)$    $[D] (IV) (i) (Q)$

($3$) The only CORRECT combination that gives two different carboxylic acids is

$[A] (IV) (iii) (Q)$ $[B] (III) (iii) (P)$ $[C] (II) (iv) (R)$ $[D] (I) (i) (S)$

Given the answer quetion ($1$) ($2$) and ($3$)