$A$ plane meets the coordinate axes at $A, B, C$ respectively such that the centroid of the $\triangle ABC$ is $(2, 3, 5)$. Then, the equation of that plane is

The image of the point $(1, 2, -1)$ on the plane containing the line $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ and the point $(0, 7, -7)$ is:

Let $\pi$ be the plane passing through the point $(3,-3,1)$ and perpendicular to the line joining the points $(3,4,-1)$ and $(2,-1,5)$. If the equation of the plane containing the points $(3,4,-1),(-1,2,5)$ and perpendicular to the plane $\pi$ is $ax+y+cz-d=0$,then $3(a+c)=$

The equation of the plane which passes through $(2, -3, 1)$ and is normal to the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$ is given by

The Cartesian equation of the plane,passing through the points $(3,1,1)$,$(1,2,3)$ and $(-1,4,2)$,is

The points $A(-1, 3, 0)$,$B(2, 2, 1)$,and $C(1, 1, 3)$ determine a plane. The distance from the plane to the point $D(5, 7, 8)$ is