When sulphur boiled with $\mathrm{NaOH}$ solution forms $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ and $\mathrm{Na}_2 \mathrm{~S}$.
$4 \mathrm{~S}+6 \mathrm{NaOH} \longrightarrow \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+2 \mathrm{Na}_2 \mathrm{~S}+3 \mathrm{H}_2 \mathrm{O}$
It is a disproportionation reaction.