Option (i) and (iii) are the answers.
They give ethyne on treatment with alcoholic $\mathrm{KOH}$.
$
\mathrm{CH}_{3} \mathrm{CHCl}_{2} \underset{\mathrm{KOH}}{\stackrel{\text { alc. }}{\longrightarrow}} \mathrm{CH} \equiv \mathrm{CH}+2 \mathrm{KCl}+2 \mathrm{H}_{2} \mathrm{O}$
$
\mathrm{Cl}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Cl} \stackrel{\text { alc. } \mathrm{KOH}}{\longrightarrow} \mathrm{CH} \equiv \mathrm{CH}+2 \mathrm{KCl}+\mathrm{H}_{2} \mathrm{O}$
On reduction with $\mathrm{Zn}$ dust in alcohol they give ethylene. $\mathrm{CH}_{3} \mathrm{CHCl}_{2}+\mathrm{Zn} \stackrel{\mathrm{CH}_{3} \mathrm{OH}}{\longrightarrow} \mathrm{CH}_{2}=\mathrm{CH}$
$
\mathrm{Cl}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Cl}+\mathrm{Zn} \stackrel{\mathrm{CH}_{3} \mathrm{OH}}{\longrightarrow} \mathrm{CH}_{2}=\mathrm{CH}_{2}$