Let the point $$P(\alpha, \beta)$$ be at a unit distance from each of the two lines $$L_{1}: 3 x-4 y+12=0$$, and $$L_{2}: 8 x+6 y+11=0$$. If $$P$$ lies below $$L_{1}$$ and above $${ }{L_{2}}$$, then $$100(\alpha+\beta)$$ is equal to :

A line, with the slope greater than one, passes through the point $$A(4,3)$$ and intersects the line $$x-y-2=0$$ at the point B. If the length of the line segment $$A B$$ is $$\frac{\sqrt{29}}{3}$$, then $$B$$ also lies on the line :

Let $$\alpha$$_{1}, $$\alpha$$_{2} ($$\alpha$$_{1} < $$\alpha$$_{2}) be the values of $$\alpha$$ fo the points ($$\alpha$$, $$-$$3), (2, 0) and (1, $$\alpha$$) to be collinear. Then the equation of the line, passing through ($$\alpha$$_{1}, $$\alpha$$_{2}) and making an angle of $${\pi \over 3}$$ with the positive direction of the x-axis, is :

The distance of the origin from the centroid of the triangle whose two sides have the equations $$x - 2y + 1 = 0$$ and $$2x - y - 1 = 0$$ and whose orthocenter is $$\left( {{7 \over 3},{7 \over 3}} \right)$$ is :