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**Hint:**Draw the line x –y = 4 and with the help of slope and distance formula we can find the value of Q. With the point Q and it is given that the line is perpendicular to x –y =4, we can find the required equation using slope point formula.

**Complete step-by-step answer:**Lets sketch the line x – y =4. Let the point (2 ,1) be P.

It is given that PQ is parallel to the x – y = 4.

Now let's draw the graph with the above details

Step 2:

Let the new point be Q(a,b)

Now from the diagram , we can see that AQ is parallel to the given line.

The slope of the given line is 1

And since AQ is parallel , their slopes are equal

Therefore slope of AQ is 1

$

\Rightarrow \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = 1 \\

\Rightarrow \dfrac{{1 - b}}{{2 - a}} = 1 \\

\Rightarrow 1 - b = 2 - a \\

\Rightarrow a - b = 1 \\

\Rightarrow a = 1 + b \\

$

Let the above equation be equation (1)

Step 3:

It is also given that P(2 , 1) is at a distance of$2\sqrt 3 $ units from Q

Distance between two points is given by

$ \Rightarrow \sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2}} $

Here we already know that the distance between P and Q is $2\sqrt 3 $ units

Therefore ,

$

\Rightarrow \sqrt {{{(2 - a)}^2} + {{(1 - b)}^2}} = 2\sqrt 3 \\

\\

$

Squaring on both sides we get,

$ \Rightarrow {(2 - a)^2} + {(1 - b)^2} = {(2\sqrt 3 )^2}$

Lets substitute equation (1) in the above equation

$

\Rightarrow {(2 - (1 + b))^2} + {(1 - b)^2} = 4*3 \\

\Rightarrow {(1 - b)^2} + {(1 - b)^2} = 12 \\

\Rightarrow 2{(1 - b)^2} = 12 \\

\Rightarrow {(1 - b)^2} = 6 \\

\Rightarrow 1 - b = \pm \sqrt 6 \\

\Rightarrow b = 1 \pm \sqrt 6 \\

$

Substitute the value of b in equation (1)

$

\Rightarrow a = b + 1 \\

\Rightarrow a = 1 \pm \sqrt 6 + 1 = 2 \pm \sqrt 6 \\

$

Step 4:

Here we have $a = 2 \pm \sqrt 6 $ and $b = 1 \pm \sqrt 6 $

Since the point Q lies in the third quadrant both the coordinates should be negative .

So $a = 2 - \sqrt 6 $and $b = 1 - \sqrt 6 $

Step 5:

Now we need to find the equation of the line passing through Q and perpendicular to x – y =4

Slope of x –y=4 is 1

Since our required equation is perpendicular to x –y=4.

The slope of our required line is -1

By using slope point formula

$(y - {y_1}) = m(x - {x_1})$

We have

$

\Rightarrow \left( {y - (1 - \sqrt 6 )} \right) = - 1\left( {x - (2 - \sqrt 6 )} \right) \\

\Rightarrow \left( {y - 1 + \sqrt 6 } \right) = - 1\left( {x - 2 + \sqrt 6 } \right) \\

\Rightarrow y - 1 + \sqrt 6 = - x + 2 - \sqrt 6 \\

\Rightarrow x + y = 2 - \sqrt 6 + 1 - \sqrt 6 \\

\Rightarrow x + y = 3 - 2\sqrt 6 \\

$

Therefore the required equation is $x + y = 3 - 2\sqrt 6 $

**The correct option is C.**

**Note:**1. Slope of parallel lines are equal.

2.Product of the slope of perpendicular lines is -1.

3.In a graph, there are four quadrants based on the sign of x and y coordinate. In the first quadrant both x and y are positive and in the second quadrant x is negative and y is positive , in the third quadrant both x and y are negative and in the fourth quadrant y is negative and x is positive.

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