The point 2 , 1 is translated parallel to the line L : x - y = 4 by 2 3 units. If the new point Q lies in…

Equation of line $QR$ perpendicular to $L:x-y=4$ is
$x+y-c=0 \dots \left(1\right)$

Distance of $QR$ from $\left(2,1\right)$ is $2\sqrt{3}$

$\Rightarrow 2\sqrt{3}=\frac{\left|2+1-c\right|}{\sqrt{2}}$

$\Rightarrow 2\sqrt{6}=\left|3-c\right|$

$\Rightarrow c-3=\pm 2\sqrt{6} \Rightarrow c=3\pm 2\sqrt{6}$

Line can be $x+y=3\pm 2\sqrt{6}$ 

$x+y=3-2\sqrt{6}$   (since $Q$ is in the third quadrant)