[b]Q13.[/b] Determine the greatest value of the sum $M=11xy+3x+2012yz$, where $x,y,z$ are non negative integers satisfying condition $x+y+z=1000.$
1. We start with the given expression \( M = 11xy + 3x + 2012yz \) and the constraint \( x + y + z = 1000 \), where \( x, y, z \) are non-negative integers.
2. To find the maximum value of \( M \), we can express \( x \) in terms of \( y \) and \( z \) using the constraint:
   \[
   x = 1000 - y - z
   \]
3. Substitute \( x = 1000 - y - z \) into \( M \):
   \[
   M = 11(1000 - y - z)y + 3(1000 - y - z) + 2012yz
   \]
   Simplify the expression:
   \[
   M = 11000y - 11y^2 - 11yz + 3000 - 3y - 3z + 2012yz
   \]
   Combine like terms:
   \[
   M = 11000y - 11y^2 + (2012 - 11)yz + 3000 - 3y - 3z
   \]
   \[
   M = 11000y - 11y^2 + 2001yz + 3000 - 3y - 3z
   \]
4. To maximize \( M \), we need to consider the values of \( y \) and \( z \). Notice that the term \( 2001yz \) will be maximized when both \( y \) and \( z \) are as large as possible.
5. By the AM-GM inequality, the product \( yz \) is maximized when \( y = z \). Given \( y + z = 1000 - x \), the maximum product occurs when \( y = z = 500 \) and \( x = 0 \).
6. Substitute \( y = 500 \) and \( z = 500 \) into \( M \):
   \[
   M = 11 \cdot 0 \cdot 500 + 3 \cdot 0 + 2012 \cdot 500 \cdot 500
   \]
   \[
   M = 0 + 0 + 2012 \cdot 250000
   \]
   \[
   M = 503000000
   \]
7. Therefore, the maximum value of \( M \) is \( 503000000 \).

The final answer is \( \boxed{503000000} \).