Let a be the length of a side of a square OABC with O being the origin. Its side OA makes an acute angle $$\alpha $$ with the positive x-axis and the equations of its diagonals are $(\sqrt{3}+1)x+(\sqrt{3}-1)y=0$ and $(\sqrt{3}-1)x-(\sqrt{3}+1)y+8\sqrt{3}=0$. Then $a$² is equal to :

A line passing through the point P($a$, 0) makes an acute angle $$\alpha $$ with the positive x-axis. Let this line be rotated about the point P through an angle $\frac{\alpha}{2}$ in the clockwise direction. If in the new position, the slope of the line is $2 - \sqrt{3}$ and its distance from the origin is $\frac{1}{\sqrt{2}}$, then the value of $3a^2 \tan^2 \alpha - 2\sqrt{3}$ is :

If the orthocenter of the triangle formed by the lines y = x + 1, y = 4x - 8 and y = mx + c is at (3, -1), then m - c is :

Let ABC be the triangle such that the equations of lines AB and AC be $3 y-x=2$ and $x+y=2$, respectively, and the points B and C lie on $x$-axis. If P is the orthocentre of the triangle ABC , then the area of the triangle PBC is equal to