$$\begin{aligned} 252 &= 3x + 7y + 14z \ge 3\sqrt[3]{3x \cdot 7y \cdot 14z} = 3\sqrt[3]{3 \cdot 7 \cdot 14 \cdot (2016 + u^2)} \\ &\ge 3\sqrt[3]{3 \cdot 7 \cdot 14 \cdot 2016} = 3\sqrt[3]{2^6 \cdot 3^3 \cdot 7^3} = 2^2 \cdot 3^2 \cdot 7 = 252. \end{aligned}$$
To not get a contradiction, we must have equality in both inequalities, hence $3x = 7y = 14z$ and $u = 0$. From the first equation of the system we finally obtain $3x = 7y = 14z = \frac{252}{3} = 84$, hence $x = 28, y = 12$ and $z = 6$.

The second equation implies $xyz \ge 2016$. Substituting the value of $x$ from the first equation here gives $(252 - 7y - 14z)yz \ge 6048$, which is equivalent to $(36 - y - 2z)yz \ge 864$. We find the maximum of function $f(y,z) = (36 - y - 2z)yz$. Fixing $z > 0$ arbitrarily, we obtain $\frac{df(y,z)}{dy} = 36z - 2yz - 2z^2 = 2z(18 - y - z)$. The partial derivative with respect to $y$ is zero, if $z = 18 - y$. As the second derivative is negative, the function $f(y,z)$ has exactly one maximum at $y = 18 - z$ for any fixed positive number $z$. Define $g(z) = f(18 - z, z) = (18 - z)^2 z$. Its derivative $g'(z) = (18 - z)(18 - 3z)$ is zero at $z = 6$ ($z = 18$ does not count since then $y = 0$). As the second derivative is negative at $z = 6$, the extremum found is a maximum again. This means that out of all partial maxima of $f(y,z)$, the one for $z = 6$ is the largest. As $g(6) = f(12,6) = 864$, the initial system of equations can be satisfied only if $y = 12$ and $z = 6$. Substituting these values into the initial system leads to $3x = 84, 72x - u^2 = 2016$. The only solution of this system is $(x, u) = (28,0)$.