Let the plane P: $$4 x-y+z=10$$ be rotated by an angle $$\frac{\pi}{2}$$ about its line of intersection with the plane $$x+y-z=4$$. If $$\alpha$$ is the distance of the point $$(2,3,-4)$$ from the new position of the plane $$\mathrm{P}$$, then $$35 \alpha$$ is equal to :

Let the line passing through the points $$\mathrm{P}(2,-1,2)$$ and $$\mathrm{Q}(5,3,4)$$ meet the plane $$x-y+z=4$$ at the point $$\mathrm{R}$$. Then the distance of the point $$\mathrm{R}$$ from the plane $$x+2 y+3 z+2=0$$ measured parallel to the line $$\frac{x-7}{2}=\frac{y+3}{2}=\frac{z-2}{1}$$ is equal to :

Let P be the plane passing through the points $$(5,3,0),(13,3,-2)$$ and $$(1,6,2)$$. For $$\alpha \in \mathbb{N}$$, if the distances of the points $$\mathrm{A}(3,4, \alpha)$$ and $$\mathrm{B}(2, \alpha, a)$$ from the plane P are 2 and 3 respectively, then the positive value of a is :

Let $$(\alpha, \beta, \gamma)$$ be the image of the point $$\mathrm{P}(2,3,5)$$ in the plane $$2 x+y-3 z=6$$. Then $$\alpha+\beta+\gamma$$ is equal to :