Solution:

First, note that the equation is equivalent to $(x-1001)+(y-601)+(z-401)=13$. Letting $x' = x-1001$, $y' = y-601$, and $z' = z-401$, we can instead count the number of solutions of $x' + y' + z' = 13$, where $x'$, $y'$, $z'$ are nonnegative integers. Now note that each solution $(x', y', z')$ to this equation corresponds to a way of choosing 13 items from a pile of objects of 3 types; the number of such choices is
$$\binom{13+3-1}{13} = \frac{15!}{13!2!} = 105$$