If the given planes $ax + by + cz + d = 0$ and $a'x + b'y + c'z + d' = 0$ are mutually perpendicular,then:

If a plane meets the coordinate axes at $A, B$ and $C$ in such a way that the centroid of $\triangle ABC$ is at the point $(1, 2, 3)$,then the equation of the plane is

The distance from a point $(1, 1, 1)$ to a variable plane $\pi$ is $12$ units and the points of intersections of the plane $\pi$ and $X, Y, Z$-axes are $A, B, C$ respectively. If the point of intersection of the planes through the points $A, B, C$ and parallel to the coordinate planes is $P$,then the equation of the locus of $P$ is

If $P$ is the point $(2, 6, 3)$,then the equation of the plane passing through $P$ and perpendicular to $OP$,where $O$ is the origin,is:

In the following case,determine the direction cosines of the normal to the plane and the distance from the origin: $x+y+z=1$

Find the vector equation of the plane which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector from the origin is $2 \hat{i}-3 \hat{j}+4 \hat{k} .$ Also find its cartesian form.