If the pair of straight line given by Ax² + 2Hxy + By² = 0, where (H²>AB), forms an equilateral triangle with the line ax + by + c = 0, then (A + 3B)(3A + B) =

Ax² + 2Hxy + By² = 0, ax + by + c = 0
Form an equilateral triangle
$\Rightarrow$Ax² + 2Hxy + By² = 0, (ax + by)² - 3(bx - ay) = 0
Represent the same lines
$\Rightarrow$A = a² - 3b², B  b² - 3a² , H = 4ab
$\therefore$(A + 3B)(3A + B) = (-8a²)(8b²) = 64a²b² = 4H²