A line $L$ has intercepts $a$ and $b$ on the coordinate axes. When the coordinate axes are rotated through an angle $\alpha$ and keeping the origin fixed, the same line $L$ has intercepts $p$ and $q$ on the new axes. Then,

Two lines $L_1$ and $L_2$ passing through the point $P(1,2)$ cut the line $x+y=4$ at a distance of $\frac{\sqrt{6}}{3}$ units from $P$. Then, the angles made by $L_1, L_2$ with positive $X$-axis are

A pair of straight lines drawn though the origin forms. an isosceles triangle right angled at the origin with the line $2 x+3 y=6$. The area (in sq units) of the triangle, so formed is

The equation of the straight line passing through the point $(3,2)$ and inclined at an angle of $60^{\circ}$ with the line $\sqrt{3} x+y=1$ is