The shortest distance between the lines $$\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}$$ and $$\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}$$ is :

Let $$R$$ be the focus of the parabola $$y^{2}=20 x$$ and the line $$y=m x+c$$ intersect the parabola at two points $$P$$ and $$Q$$.

Let the point $$G(10,10)$$ be the centroid of the triangle $$P Q R$$. If $$c-m=6$$, then $$(P Q)^{2}$$ is :

The area of the region $$\left\{(x, y): x^{2} \leq y \leq 8-x^{2}, y \leq 7\right\}$$ is :

Let $$C(\alpha, \beta)$$ be the circumcenter of the triangle formed by the lines

$$4 x+3 y=69$$

$$4 y-3 x=17$$, and

$$x+7 y=61$$.

Then $$(\alpha-\beta)^{2}+\alpha+\beta$$ is equal to :