Prime Triplet

Answer would be 1. As $n$ , $n+2$ , $n+4$ are all prime so these numbers are are divisible by $1$ and itself. Now consider the product of all these numbers i.e $n*n+2*n+4$ $\equiv$ $n*n+1*n+2$ ( $mod 3$ ) $\equiv$ $0$ ( $mod 3$ ) $\implies$ 3 divides the product which forces n to be $3$ otherwise product would have factors $2,3,6$ which contradicts that product has only $n,n+2,n+4$ as factors.

All prime numbers (greater than or equal to $5$ ) are of the form $6q +1$ or $6q + 5$ , where $q$ is greater or equal than $1$ .
{Since $6q +1$ and $6q + 5$ both are only divisible by 1 and the number itself)}
Let $n$ be a prime number greater or equal to $5$ { $q$ greater than or equal to $1$ }.
$Case 1 :$
$n = 6q + 1$ (a prime number);
$n + 2 = 6q + 3$ ( can never be a prime number);
$n + 4 = 6q + 5$ ( can be a prime number);
$Case 2 :$
$n = 6q + 5$ ( a prime number);
$n + 2 = 6q + 7 = 6(q + 1) + 1$ ( can be a prime number);
$n + 4 = 6q + 9$ (not a prime number as it is divisible by $3$ );

Therefore for all cases $n$ greater than $7$ no such triple exists. So the only possible solution is $(3,5,7)$ . Answer is $\boxed{1}$ .

$Q.E.D$

One of the numbers n, n+2, n+4 is divisible by 3 for all integer n. (since n has to be of the form 3m or 3m-1 or 3m+1 for integer m) . Therefore the only possibility is the triplet 3, 5, 7