The image point of $(1, 3, 4)$ in the plane $2x - y + z + 3 = 0$ is

The equation of the plane passing through $(1, -1, 2)$ and perpendicular to the planes $x + 2y - 2z = 4$ and $3x + 2y + z = 6$ is:

If $(2, -1, 3)$ is the foot of the perpendicular drawn from the origin $(0, 0, 0)$ to a plane,then the equation of that plane is:

$A$ plane $\pi$ passes through the points $(5,1,2)$, $(3,-4,6)$, and $(7,0,-1)$. If $p$ is the perpendicular distance from the origin to the plane $\pi$ and $l, m, n$ are the direction cosines of a normal to the plane $\pi$, then $|3l+2m+5n|=$

$A$ variable plane is at a constant distance $h$ from the origin and meets the coordinate axes in $A, B, C$. The locus of the centroid of $\triangle ABC$ is

If the line drawn from the point $(-2, -1, -3)$ meets a plane at a right angle at the point $(1, -3, 3)$,find the equation of the plane.