is ${(a+b)! \over (a+b)(c!)}-1$= prime number if a,b and c are primes? if 2a=b+c and b>a&c

Question

${(a+b)! \over (a+b)(c!)}-1$ firstly $a,b$ and $c$ need to be prime numbers then the sum of $b+c = 2a$ or double a secondly $b>c$ for example if $a=5, b=7, c=3$ then that gives me $6652799$ which is a prime. However I'm asking if you can prove or disprove that for every $a,b$ and $c$ that make the requirement make a prime number.

Answer 1

No, take $a=11,b=19,c=3$. You get $$1473626998956616992423935999999=127*4637*47947*33012437*1580912118659$$