Suppose  that  $x,  y,  z$  are  three  distinct  prime  numbers  such  that  $x  +  y  +  z  =  49$.  Find  the maximum  possible  value  for  the  product  $xyz$.

$\text{(A) } 615 \quad
\text{(B) } 1295 \quad
\text{(C) } 2387 \quad
\text{(D) } 1772 \quad
\text{(E) } 715 \quad
\text{(F) } 442 \quad
\text{(G) } 1479 \quad \\
\text{(H) } 2639 \quad
\text{(I) } 3059 \quad
\text{(J) } 3821 \quad
\text{(K) } 3145 \quad
\text{(L) } 1715 \quad
\text{(M) } \text{none of the above} \quad $
1. We are given that \(x, y, z\) are three distinct prime numbers such that \(x + y + z = 49\). We need to find the maximum possible value for the product \(xyz\).

2. To maximize the product \(xyz\), we should choose the prime numbers \(x, y, z\) such that they are as close to each other as possible. This is because the product of numbers that are close to each other is generally larger than the product of numbers that are far apart.

3. Let's list the prime numbers less than 49: \(2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47\).

4. We need to find three distinct primes from this list that sum to 49. We can start by trying combinations of primes that are close to each other:
   - \(13 + 17 + 19 = 49\)
   - \(11 + 17 + 21\) (21 is not a prime)
   - \(11 + 19 + 19\) (19 is repeated)
   - \(7 + 19 + 23 = 49\)
   - \(5 + 23 + 21\) (21 is not a prime)
   - \(3 + 23 + 23\) (23 is repeated)
   - \(2 + 23 + 24\) (24 is not a prime)

5. From the above combinations, the valid sets of primes that sum to 49 are:
   - \(13, 17, 19\)
   - \(7, 19, 23\)

6. Now, we calculate the product for each valid set:
   - For \(13, 17, 19\):
     \[
     13 \times 17 \times 19 = 4199
     \]
   - For \(7, 19, 23\):
     \[
     7 \times 19 \times 23 = 3059
     \]

7. Comparing the products, \(4199\) is greater than \(3059\).

8. Therefore, the maximum possible value for the product \(xyz\) is \(4199\).

The final answer is \(\boxed{4199}\).