For how many primes $p$ less than $15$, there exists integer triples $(m,n,k)$ such that
\[
\begin{array}{rcl}
m+n+k &\equiv& 0 \pmod p \\
mn+mk+nk &\equiv& 1 \pmod p \\
mnk &\equiv& 2 \pmod p.
\end{array}
\]

$ 
\textbf{(A)}\ 2
\qquad\textbf{(B)}\ 3
\qquad\textbf{(C)}\ 4
\qquad\textbf{(D)}\ 5
\qquad\textbf{(E)}\ 6
$
To solve the problem, we need to find the number of primes \( p \) less than 15 for which there exist integer triples \((m, n, k)\) satisfying the given congruences:
\[
\begin{array}{rcl}
m+n+k &\equiv& 0 \pmod{p} \\
mn+mk+nk &\equiv& 1 \pmod{p} \\
mnk &\equiv& 2 \pmod{p}.
\end{array}
\]

1. **Formulating the Polynomial:**
   The given system of congruences can be interpreted as the roots of a polynomial. Specifically, if \( m, n, k \) are roots of a polynomial \( P(x) \) modulo \( p \), then:
   \[
   P(x) = (x - m)(x - n)(x - k).
   \]
   Expanding this, we get:
   \[
   P(x) = x^3 - (m+n+k)x^2 + (mn+mk+nk)x - mnk.
   \]
   Given the conditions:
   \[
   m+n+k \equiv 0 \pmod{p}, \quad mn+mk+nk \equiv 1 \pmod{p}, \quad mnk \equiv 2 \pmod{p},
   \]
   we can rewrite the polynomial as:
   \[
   P(x) = x^3 + x - 2 \pmod{p}.
   \]

2. **Factorization in \(\mathbb{Z}_p[x]\):**
   We need \( x^3 + x - 2 \) to factor into three linear factors in \(\mathbb{Z}_p[x]\). This means that \( x^3 + x - 2 \) must have three distinct roots modulo \( p \).

3. **Checking for Quadratic Residue:**
   We can factor \( x^3 + x - 2 \) as:
   \[
   x^3 + x - 2 = (x - 1)(x^2 + x + 2).
   \]
   For this factorization to hold in \(\mathbb{Z}_p[x]\), \( x^2 + x + 2 \) must have a root modulo \( p \). This is equivalent to checking if the discriminant of \( x^2 + x + 2 \) is a quadratic residue modulo \( p \).

   The discriminant of \( x^2 + x + 2 \) is:
   \[
   \Delta = 1^2 - 4 \cdot 1 \cdot 2 = 1 - 8 = -7.
   \]
   Therefore, we need \(-7\) to be a quadratic residue modulo \( p \).

4. **Checking Primes Less Than 15:**
   We check each prime \( p \) less than 15 to see if \(-7\) is a quadratic residue modulo \( p \).

   - For \( p = 2 \):
     \[
     -7 \equiv 1 \pmod{2}, \quad 1 \text{ is a quadratic residue modulo } 2.
     \]
   - For \( p = 3 \):
     \[
     -7 \equiv -1 \equiv 2 \pmod{3}, \quad 2 \text{ is not a quadratic residue modulo } 3.
     \]
   - For \( p = 5 \):
     \[
     -7 \equiv 3 \pmod{5}, \quad 3 \text{ is not a quadratic residue modulo } 5.
     \]
   - For \( p = 7 \):
     \[
     -7 \equiv 0 \pmod{7}, \quad 0 \text{ is a quadratic residue modulo } 7.
     \]
   - For \( p = 11 \):
     \[
     -7 \equiv 4 \pmod{11}, \quad 4 \text{ is a quadratic residue modulo } 11.
     \]
   - For \( p = 13 \):
     \[
     -7 \equiv 6 \pmod{13}, \quad 6 \text{ is not a quadratic residue modulo } 13.
     \]

   Therefore, the primes \( p \) less than 15 for which \(-7\) is a quadratic residue are \( 2, 7, \) and \( 11 \).

Conclusion:
The number of primes \( p \) less than 15 for which there exist integer triples \((m, n, k)\) satisfying the given conditions is \( \boxed{3} \).