3. Given positive integers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+z^{2}=2(y z+1), \\
x+y+z=4032 .
\end{array}\right.
$$

Then the value of $x^{2} y+z$ is ( ).
(A) 2015
(B) 2016
(C) 4031
(D) 4033
3. C.

Since $x^{2}+(y-z)^{2}=2$, and $x$ is a positive integer, $y-z$ is an integer, so, $x=1$.
Therefore, $y+z=4032-x=4031$.