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The point $(2, 1)$ is translated parallel to the line $L : x−y=4$ by $2 \sqrt 3$ units. If the new point Q lies in the third quadrant, then the equation of the line passing through Q and perpendicular to L is :

$\begin{array}{1 1} (1) x+y=2 - \sqrt {6} \\ (2) x+y=3- 3 \sqrt {6} \\(3) x+y=3 - 2\sqrt {6} \\ (4) 2x+2y= 1-\sqrt{ 6} \end{array} $

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