JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Sets Relations and Functions

The complement of $B$ consists of the subsets that contain $1$ but not $2$; the remaining five elements $\{3,4,5,6,7\}$ are then free, giving $2^5=32$ such subsets, so $|B|=2^7-32=96$. To build $B\cup C$ one adds the members of $C$ lying outside $B$, that is subsets $T$ containing $1$, not containing $2$, with prime element-sum. Writing $T=\{1\}\cup S$ with $S\subseteq \{3,4,5,6,7\}$, the sum is $1+\sigma (S)$, so $1+\sigma (S)$ must be prime. Listing the attainable values of $\sigma (S)$ with their multiplicities, the prime totals occur for $\sigma (S)=4,6,10,12,16,18,22$, contributing $1+1+2+2+2+2+1=11$ subsets. Therefore $|B\cup C|=96+11=107$.