The perpendicular distance of the origin from the plane $$x-3 y+4 z-6=0$$ is

Two lines $$\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}$$ and $$\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4} \quad$$ intersect at the point R. Then reflection of $$\mathrm{R}$$ in the $$x y$$-plane has co-ordinates

The line $$\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$$ lies in the plane $$x+3 y-\alpha z+\beta=0$$, then the value of $$\alpha^2+\alpha \beta+\beta^2$$ is

Let $$\mathrm{P}$$ be a plane passing through the points $$(2,1,0),(4,1,1)$$ and $$(5,0,1)$$ and $$R$$ be the point $$(2,1,6)$$. Then image of $$R$$ in the plane $$P$$ is