The d.r.s. of the normal to the plane passing through the origin and the line of intersection of the planes $$x+2 y+3 z=4$$ and $$4 x+3 y+2 z=1$$ are

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 value of $$\alpha \beta$$ is

If the points $$P(4,5, x), Q(3, y, 4)$$ and $$R(5,8,0)$$ are collinear, then the value of $$x+y$$ is

A line drawn from a point $$A(-2,-2,3)$$ and parallel to the line $$\frac{x}{-2}=\frac{y}{2}=\frac{z}{-1}$$ meets the $$\mathrm{YOZ}$$ plane in point $$\mathrm{P}$$, then the co-ordinates of the point $$\mathrm{P}$$ are