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 :

If the ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ meets the line $$\frac{x}{7}+\frac{y}{2 \sqrt{6}}=1$$ on the $$x$$-axis and the line $$\frac{x}{7}-\frac{y}{2 \sqrt{6}}=1$$ on the $$y$$-axis, then the eccentricity of the ellipse is :

Let the foci of the ellipse $$\frac{x^{2}}{16}+\frac{y^{2}}{7}=1$$ and the hyperbola $$\frac{x^{2}}{144}-\frac{y^{2}}{\alpha}=\frac{1}{25}$$ coincide. Then the length of the latus rectum of the hyperbola is :

The shortest distance between the lines $$\frac{x+7}{-6}=\frac{y-6}{7}=z$$ and $$\frac{7-x}{2}=y-2=z-6$$ is :