The plane passing through the points $(1,1,1)$,$(1,-1,1)$ and $(-7,-3,-5)$ is

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

$A$ plane passing through $(-1, 2, 3)$ and whose normal makes equal angles with the coordinate axes is

The distance between the two parallel planes $2x + y + 2z = 8$ and $4x + 2y + 4z + 5 = 0$ is: (in $/2$)

In three-dimensional space,the equation $3y + 4z = 0$ represents:

Let the image of the point $P(1, 2, 6)$ in the plane passing through the points $A(1, 2, 0)$,$B(1, 4, 1)$,and $C(0, 5, 1)$ be $Q(\alpha, \beta, \gamma)$. Then $(\alpha^2 + \beta^2 + \gamma^2)$ is equal to :