$\large{\begin{cases} 2{ x }^{ 2 }-yz\left( 2-\sqrt{2}+\sqrt { 6 } \right) =\left( y-z \right) ^{ 2 }\times 2 \\ z^{ 2 }+xy(\sqrt { 3 } +2)={ (x+y) }^{ 2 } \\ \frac { y+z }{ x } =\frac { x-az }{ y-z } \end{cases}}$

For some positive real numbers $x, y, z$ satisfy the system of equations above. For some particular value of $a$ ( $a \in \mathbb{R^{+}}$ ) and the ratio $\dfrac{z}{y}$ ( let's say the ratio be $m$ ), there exist infinitely many solutions for the above given system in $(x,y,z)$ . Find

$\left\lfloor 1000a \right\rfloor +\left\lceil 1000m \right\rceil$

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The answer is 1768.

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