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 the third quadrant, then the equation of the line passing through Q and perpendicular to L is:

Let (2,1) be point A. (2,1) is at a distance of 2√3 units from point Q(a,b).
The direction ratios of line L: x - y = 4 are (1, -1).
The direction ratios of the line parallel to L are (1, -1).
Let Q = (2 + k, 1 - k), where k is a scalar.
The distance between A(2,1) and Q(2+k, 1-k) is 2√3.
√((2+k-2)² + (1-k-1)²) = 2√3
√(k² + k²) = 2√3
√(2k²) = 2√3
|k√2| = 2√3
k = ±2√3/√2 = ±√6
Since Q lies in the third quadrant, both coordinates must be negative.
Thus, k = -√6
Q = (2 - √6, 1 + √6)
Slope of L = 1
Slope of line perpendicular to L = -1
Equation of line passing through (2 - √6, 1 + √6) with slope -1 is:
y - (1 + √6) = -1(x - (2 - √6))
y - 1 - √6 = -x + 2 - √6
x + y = 3