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

The given point is (2, 1). We need to translate this point parallel to the line L: $$x - y = 4$$ by a distance of $$2\sqrt{3}$$ units. The new point Q lies in the third quadrant, and we must find the equation of the line passing through Q and perpendicular to L.

First, find the direction vector of line L. The line $$x - y = 4$$ can be rewritten as $$y = x - 4$$, so its slope is 1. Therefore, the direction vector is proportional to (1, 1). The magnitude of this vector is $$\sqrt{1^2 + 1^2} = \sqrt{2}$$. The unit vector in the direction of (1, 1) is $$\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$$.

The translation distance is $$2\sqrt{3}$$ units, so the displacement vector can be either $$2\sqrt{3} \times \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$$ or $$2\sqrt{3} \times \left( -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right)$$. Simplifying:

$$2\sqrt{3} \times \frac{1}{\sqrt{2}} = \frac{2\sqrt{3}}{\sqrt{2}} = \sqrt{\frac{4 \times 3}{2}} = \sqrt{6}$$

So the displacement vectors are $$(\sqrt{6}, \sqrt{6})$$ and $$(-\sqrt{6}, -\sqrt{6})$$.

The two possible new points are:

Option 1: $$(2 + \sqrt{6}, 1 + \sqrt{6})$$

Option 2: $$(2 - \sqrt{6}, 1 - \sqrt{6})$$

Since Q lies in the third quadrant, both coordinates must be negative. Approximating $$\sqrt{6} \approx 2.45$$:

For Option 1: $$2 + 2.45 = 4.45 > 0$$ and $$1 + 2.45 = 3.45 > 0$$, so it is in the first quadrant.

For Option 2: $$2 - 2.45 = -0.45 < 0$$ and $$1 - 2.45 = -1.45 < 0$$, so it is in the third quadrant.

Thus, Q is $$(2 - \sqrt{6}, 1 - \sqrt{6})$$.

Now, find the line perpendicular to L passing through Q. The slope of L is 1, so the slope of a perpendicular line is the negative reciprocal, which is $$-1$$.

The equation of a line with slope $$-1$$ passing through $$(x_0, y_0)$$ is $$y - y_0 = m(x - x_0)$$. Substituting $$m = -1$$, $$x_0 = 2 - \sqrt{6}$$, and $$y_0 = 1 - \sqrt{6}$$:

$$y - (1 - \sqrt{6}) = -1 \times (x - (2 - \sqrt{6}))$$

Simplify the right side:

$$y - 1 + \sqrt{6} = - (x - 2 + \sqrt{6})$$

$$y - 1 + \sqrt{6} = -x + 2 - \sqrt{6}$$

Bring all terms to the left side:

$$y - 1 + \sqrt{6} + x - 2 + \sqrt{6} = 0$$

Combine like terms:

$$x + y + (-\1 - 2) + (\sqrt{6} + \sqrt{6}) = 0$$

$$x + y - 3 + 2\sqrt{6} = 0$$

Rearrange to standard form:

$$x + y = 3 - 2\sqrt{6}$$

Comparing with the options:

A. $$x + y = 2 - \sqrt{6}$$

B. $$2x + 2y = 1 - \sqrt{6}$$ which simplifies to $$x + y = \frac{1 - \sqrt{6}}{2}$$

C. $$x + y = 3 - 3\sqrt{6}$$

D. $$x + y = 3 - 2\sqrt{6}$$

The equation $$x + y = 3 - 2\sqrt{6}$$ matches option D.

Hence, the correct answer is Option D.