The sequence of numbers 49297533, 49297534, and 49297535 is notable, because the prime factorizations of these numbers are each of the form $a^1 \cdot b^1 \cdot c^1 \cdot d^1 \cdot e^1$, and their representations in this form include the first 11 primes:
$$ 3\cdot 19\cdot 23\cdot 31\cdot 1213 = 49,297,533$$ $$ 2\cdot 11\cdot 13\cdot 97\cdot 1777 = 49,297,534$$ $$ 5\cdot 7\cdot 17\cdot 29\cdot 2857 = 49,297,535$$
What is the lowest sequence of numbers $x, x+1, x+2$ such that all 3 numbers are of this "five prime factors" form, and their representations, taken together, include the first 12 primes, $2,3,5,7,11,13,17,19,23,29,31,37$?
Note that 12 would be the largest number of minimal primes one could have in this configuration, since there is a total of 15 primes needed, meaning that with 13 minimal primes included, one of the products would be much smaller than the others.
POSTSCRIPT: I found the smallest group of cardinality 4 (using the 9 smallest primes) reasonably quickly: $$ 11\cdot 13\cdot 17\cdot 9463= 23,004,553 $$ $$ 2\cdot 19\cdot 23\cdot 26321= 23,004,554 $$ $$ 3\cdot 5\cdot 7\cdot 219091 =23,004,555 $$
and for cardinality 3: $$ 3\cdot 11\cdot 113 = 3729 $$ $$ 2\cdot 5\cdot 373 = 3730 $$ $$ 7\cdot 13\cdot 41 = 3731 $$