How many interger tuples $(x,y,z)$ are there satisfying $0\leq x,y,z < 2011$, $xy+yz+zx \equiv 0 \pmod{2011}$, and $x+y+z \equiv 0 \pmod{2011}$ ?

$\textbf{(A)}\ 2010 \qquad\textbf{(B)}\ 2011 \qquad\textbf{(C)}\ 2012 \qquad\textbf{(D)}\ 4021 \qquad\textbf{(E)}\ 4023$
1. **Case 1: One of \(x\), \(y\), or \(z\) is zero.**

   If one of \(x\), \(y\), or \(z\) is zero, say \(x = 0\), then the equations become:
   \[
   y + z \equiv 0 \pmod{2011}
   \]
   \[
   yz \equiv 0 \pmod{2011}
   \]
   Since \(y\) and \(z\) must be integers between 0 and 2010, the only solution is \(y = 0\) and \(z = 0\). This gives us one solution: \((0, 0, 0)\).

2. **Case 2: \(x, y, z \neq 0\).**

   Given the conditions:
   \[
   x + y + z \equiv 0 \pmod{2011}
   \]
   \[
   xy + yz + zx \equiv 0 \pmod{2011}
   \]
   From \(x + y + z \equiv 0 \pmod{2011}\), we can express \(z\) as:
   \[
   z \equiv -x - y \pmod{2011}
   \]
   Substituting \(z\) into the second equation:
   \[
   xy + y(-x - y) + x(-x - y) \equiv 0 \pmod{2011}
   \]
   Simplifying:
   \[
   xy - xy - y^2 - x^2 - xy \equiv 0 \pmod{2011}
   \]
   \[
   -y^2 - x^2 - xy \equiv 0 \pmod{2011}
   \]
   \[
   x^2 + xy + y^2 \equiv 0 \pmod{2011}
   \]
   Rearranging:
   \[
   (2y + x)^2 \equiv -3x^2 \pmod{2011}
   \]
   Since \(2011 \equiv 1 \pmod{3}\), \(-3x^2\) is always a quadratic residue modulo 2011. For each \(x\), there are 2 possible values for \(y\) that satisfy the equation, and then \(z\) is determined uniquely by \(z \equiv -x - y \pmod{2011}\).

   Therefore, for each \(x\) (where \(x \neq 0\)), there are 2 values of \(y\) and 1 value of \(z\). Since \(x\) can take any value from 1 to 2010 (2010 values), we have:
   \[
   2010 \times 2 = 4020
   \]
   solutions for this case.

3. **Total Solutions:**

   Adding the solutions from both cases:
   \[
   4020 + 1 = 4021
   \]

The final answer is \(\boxed{4021}\)