1. Interpret the sets

Let the universe be the power set of $A$, where $A=\{1,2,3,4,5,6,7\}.$ So total number of subsets is $2^7=128.$

We are given:

• $B=\{T\subseteq A: \text{either }1\notin T\text{ or }2\in T\}$
• $C=\{T\subseteq A: \text{sum of elements of }T\text{ is prime}\}$

We need $|B\cup C|$.

2. Find $|B|$

The condition for $B$ is: $1\notin T \quad \text{or} \quad 2\in T.$ This fails only when $1\in T \quad \text{and} \quad 2\notin T.$

So let us count the complement of $B$ in $\mathcal P(A)$:

• $1$ must be included,
• $2$ must be excluded,
• the remaining elements $3,4,5,6,7$ are arbitrary.

Hence number of such subsets is $2^5=32.$ Therefore, $|B|=128-32=96.$

3. Find $B^c$

Since we need $|B\cup C|$, it is convenient to use $|B\cup C|=|B|+|C\cap B^c|.$

Now $B^c=\{T\subseteq A:1\in T,\ 2\notin T\}.$ So we only need to count those subsets in $B^c$ whose sum is prime.

Every such subset has the form $T=\{1\}\cup S,$ where $S\subseteq \{3,4,5,6,7\}.$ Thus the sum of elements of $T$ is $1+\text{sum}(S).$

4. List all possible sums from $\{3,4,5,6,7\}$

We need $1+\text{sum}(S)$ to be prime.

Let us enumerate all subsets $S\subseteq\{3,4,5,6,7\}$ and check whether $1+\text{sum}(S)$ is prime.

There are $2^5=32$ such subsets.

(i) Empty subset

• $S=\varnothing$, sum $=0$, so total $=1$ (not prime)

(ii) One-element subsets

• $\{3\}$: total $1+3=4$ (not prime)
• $\{4\}$: total $1+4=5$ (prime)
• $\{5\}$: total $1+5=6$ (not prime)
• $\{6\}$: total $1+6=7$ (prime)
• $\{7\}$: total $1+7=8$ (not prime)

Prime totals here: $2$ subsets.

(iii) Two-element subsets

• $\{3,4\}$: total $1+7=8$ (not prime)
• $\{3,5\}$: total $1+8=9$ (not prime)
• $\{3,6\}$: total $1+9=10$ (not prime)
• $\{3,7\}$: total $1+10=11$ (prime)
• $\{4,5\}$: total $1+9=10$ (not prime)
• $\{4,6\}$: total $1+10=11$ (prime)
• $\{4,7\}$: total $1+11=12$ (not prime)
• $\{5,6\}$: total $1+11=12$ (not prime)
• $\{5,7\}$: total $1+12=13$ (prime)
• $\{6,7\}$: total $1+13=14$ (not prime)

Prime totals here: $3$ subsets.

(iv) Three-element subsets

• $\{3,4,5\}$: total $1+12=13$ (prime)
• $\{3,4,6\}$: total $1+13=14$ (not prime)
• $\{3,4,7\}$: total $1+14=15$ (not prime)
• $\{3,5,6\}$: total $1+14=15$ (not prime)
• $\{3,5,7\}$: total $1+15=16$ (not prime)
• $\{3,6,7\}$: total $1+16=17$ (prime)
• $\{4,5,6\}$: total $1+15=16$ (not prime)
• $\{4,5,7\}$: total $1+16=17$ (prime)
• $\{4,6,7\}$: total $1+17=18$ (not prime)
• $\{5,6,7\}$: total $1+18=19$ (prime)

Prime totals here: $4$ subsets.

(v) Four-element subsets

• $\{3,4,5,6\}$: total $1+18=19$ (prime)
• $\{3,4,5,7\}$: total $1+19=20$ (not prime)
• $\{3,4,6,7\}$: total $1+20=21$ (not prime)
• $\{3,5,6,7\}$: total $1+21=22$ (not prime)
• $\{4,5,6,7\}$: total $1+22=23$ (prime)

Prime totals here: $2$ subsets.

(vi) Five-element subset

• $\{3,4,5,6,7\}$: total $1+25=26$ (not prime)

Prime totals here: $0$ subsets.

So, $|C\cap B^c|=2+3+4+2=11.$

5. Compute $|B\cup C|$

Since every subset in $B$ is already counted, and among the $32$ subsets in $B^c$ exactly $11$ belong to $C$, $|B\cup C|=|B|+|C\cap B^c|=96+11=107.$

6. Final answer

$\boxed{107}$

This matches the stored correct answer.