Answer:
Let a be the smaller of the two primes.
Now the middle of 2 primes is always even. So the middle number a+1 is divisible by 2.
Next the smaller of the primes when divided by 3 can have remainders 1,2.
1 is ruled out as possible remainder because then the remainder of a+2, the bigger of the primes would be (a+2) mod 3=(1+2) mod 3=3 mod 3=0,a contradiction since a prime number(except 3) when divided by 3 cannot have 0 as remainder.
So 2 is the only possible remainder of a. So the remainder of the bigger of the two primes when divided by 3 is (a+2) mod 3= (2+2) mod 3=4 mod 3=1.
This implies the middle number must have remainder (a+1)mod 3=(2+1)mod 3=3 mod 3=0. So the middle number is divisible by 3 also.
Hence a+1 is divisible by both 3 and 2 and since 2 and 3 have no common factors, so the middle number is divisible by 6