Equation of Plane

Find the equation of a plane, given that the foot of perpendicular drawn to the plane from origin is $(2, 1, 2).$

The coordinates of the foot of the perpendicular drawn from the origin to the plane $5y+8=0$ are

The coordinates of the foot of the perpendicular drawn from the origin to the plane $2x+3y+4z-12=0$ are

The direction cosines of the normal to the plane $3x-6y+2z=7$ are

The equation of the plane which passes through the point $(2,-3,7)$and makes equal intercepts on the coordinate axes is

The equation of the plane whose intercepts on the coordinate axes are $2,-4$ and $5$ respectively is

The equation of the plane passing through the group of points $A(2, 2, -1), B(3, 4, 2)$ and $C(7, 0, 6)$ is

The image of the point $A(1,2,3)$ relative to the plane $\pi$ is $B(3,6,-1),$ the equation of plane $\pi$ is......

The equation of the plane containing the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{4}$ is 

The equation of the plane passing through the point ($\left(1,1,2\right)$ having $2,3,2$ as direction ratios of normal to the plane is 

The intercepts cut off by the plane $2x+y-z=5$ are

The coordinate of the foot of a perpendicular drawn from the origin to the plane are $\left(2,3,1\right)$, Find the vector equation of the plane.

If the foot of the perpendicular drawn from the origin to a plane is $(1, 2, 3),$ then a point on that plane is

The equation of the plane whose intercepts on $x, y, z$ axes are $1, 2, 4$, respectively is

The image of the point with position vector $(\hat{i}+3\hat{j}+4\hat{k}),$ in the plane $\overrightarrow{r}·(2\hat{i}-\hat{j}+\hat{k})+3=0$ is

If the equation of the plane through the straight line $\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z}{5}$ and perpendicular to the plane $x-y+z+2=0$ is $a x-b y+c z+4=0,$ then the value of ${10}^{3}a+{10}^{2}b+10c$ is equal to

The length of the perpendicular drawn from the point $\left(2,1,4\right)$ to the plane containing the lines $\overrightarrow{r}=\left(\hat{\mathrm{i}}+\hat{\mathrm{j}}\right)+\mathrm{\lambda }\left(\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\hat{\mathrm{k}}\right)$ and
$\overrightarrow{r}=\left(\hat{\mathrm{i}}+\hat{\mathrm{j}}\right)+\mathrm{\mu }\left(-\hat{\mathrm{i}}+\hat{\mathrm{j}}-2\hat{\mathrm{k}}\right)$ is :

A plane bisects the line segment joining the points $\left(1, 2, 3\right)$ and $\left(-3, 4, 5\right)$ at right angles. Then this plane also passes through the point