Plane and a Point

For all $d, 0<d<1,$ which one of the following points is the reflection of the point $\left(d,2d,3d\right)$ in the plane passing through the points $\left(1,0,0\right), \left(0,1,0\right)$ and $\left(0,0,1\right)$?

A variable plane is at a constant distance $p$ from the origin $O$ and meets the set of rectangular axes $O{X}_i\left(i=1,2,3\right)$ at point $A_i\left(i=1,2,3\right)$, respectively. If planes are drawn through $A_1, A_2 \text{and} A_3$, which are parallel to the coordinate planes, then the locus of their point of intersection is

Find the distance of the point $P(3, 4, 5)$ from $z$-axis.

The distance of point $(3,-2,1)$ to plane $3x-4y+12z=3$ measured in the direction of line $\frac{x-3}{3}=\frac{y+2}{-4}=\frac{z-1}{12}$ is

Foot of perpendicular from $\left(-1,3,4\right)$ on the plane $x-2y=0,$ is

Find the distance[Units] between plane $2x+4y-4z=6$ and point $M\left(0,3,6\right)$.

If the distance of the point $\hat{i}+2\hat{j}-\hat{k}$ from the plane $\overrightarrow{r}.\left(\hat{i}-2\hat{j}+4\hat{k}\right)=10$ is $\frac{a}{\sqrt{21}}$, then $a=$

The distance of the point $(1,2,-1)$ from the plane $x-2y+4z-10=0$ is

The equation of the plane passing through the points $\left(2, 3, 1\right), \left(4, -5, 3\right)$ and parallel to $y$-axis is

The foot of the perpendicular drawn from the origin to the plane $x+y+3z-4=0$ is

Let $\mathrm{A}$ be the foot of the perpendicular from the origin to the plane $\mathrm{x}-2\mathrm{y}+2\mathrm{z}+6=0$ and $\mathrm{B}$$\left(0,\mathrm{}-1,\mathrm{}-4\right)$ be a point on the plane. Then, the length of $\mathrm{A}\mathrm{B}$ is

Show that the points $\stackrel{⏜}{i}-\stackrel{⏜}{j}+3\stackrel{⏜}{k}$ and $3\stackrel{⏜}{i}+4\stackrel{⏜}{j}+3\hat{k}$ are equidistant from the plane $\overrightarrow{r}(5\stackrel{⏜}{i}+2\hat{j}-7\hat{k})+8=0$

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

A plane is passing through the points $\left(\mathrm{2,} 2, 1\right)$ and $\left(\mathrm{9,} 3, 6\right)$ and perpendicular to the plane

$2x+6y+6z-1=0$ .Find the equation of required plane.

Find the distance between two parallel planes $2x + y + 2z = 8$ and $4x + 2y + 4z + 5 = 0$.

Equation of the image of the line $\frac{x-2}{1}=\frac{y-1}{2}=\frac{z}{3}$ in the plane $2x+y+z=5$ is

A plane passing through a point $(1,2,-4)$ having normal perpendicular to two below lines

$\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}$
and
$\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$

The perpendicular distance of the point $(0,0,0)$ from plane is

The value of $i+j+k$, if a point $Y$ whose coordinates $(i, j, k)$ is the image of point $X(-6,2,3)$ in the plane $3x-4y+5z-9=0$ is:

Let $\mathrm{P}(a, b, c)$ be any point on the plane. $x+2y+3z-7=0.$ Find the least value of $a^2+b^2+c^2.$