Angle Between a Line and a Plane

The angle between the line $\frac{x-2}{1}=\frac{y+3}{-2}=\frac{z+4}{-3}$ and the plane $2x-3y+z=5$ is

The angle between the line $\overrightarrow{r}=i+2j-k+\lambda \left(i-j+k\right)$ and normal to the plane $\overrightarrow{r}·(2\hat{i}-j+k)=4$ is

The value of $m$ so that the line $\overrightarrow{r}=i+2k+\lambda (2i-mj-3k)$ is parallel to the plane $\overrightarrow{r}·\left(mi+3j+k\right)=4$ is

The ratio in which the plane $\overrightarrow{r}·(i-2j+3k)=17$ divides the line joining the points $-2i+4j+7k$ and $3i-5j+8k$ is

The line $\frac{x-1}{1}=\frac{y+3}{3}=\frac{z-2}{2}$ meets the plane $3x-2y+z=7$ at the point

The line $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4}$ meets the plane $2x+4y-z=3$ at the point

If for $a>0$, the feet of perpendiculars from the points $A\left(a,-2a,3\right)$ and $B\left(0,4,5\right)$ on the plane $lx+my+nz=0$ are points $C\left(0,-a,-1\right)$ and $D$ respectively, then the length of line segment $CD$ is equal to :

The distance of the point $\left(1,1,9\right)$ from the point of intersection of the line $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}$ and the plane $x+y+z=17$ is:

If the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3 y-\alpha z+\beta =0$ then $(\alpha ,\beta )$ is

The angle between the line $\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+2\hat{\mathrm{j}}-3\hat{\mathrm{k}})+\mathrm{t}(2\hat{\mathrm{i}}+\hat{\mathrm{j}}-2\hat{\mathrm{k}})$ and the plane$\overrightarrow{\mathrm{r}}·(\hat{\mathbf{i}}+\hat{\mathrm{j}})+4=0$ is

The distance of point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ with the plane $x-y+z=5$ from the point with position vector $\hat{\mathrm{i}}-2\hat{\mathrm{j}}+3\hat{\mathrm{k}}$ is

The distance of the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane $x-y+z=5$ from the point $\left(-1,-5,-10\right)$, is

The point of intersection of the line $\frac{x-1}{3}=\frac{y+2}{4}=\frac{z-3}{-2}$ and plane $2x-y+3z-1=0$, is

The distance of the point $\left(1,0,2\right)$ from the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane $x-y+z=16$ is

The distance of the point $(3,4,5)$ from the point of intersection of the line $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}$ and plane $x+y+z=2$ is

A line $L$ is passing through the point $A$ whose position vector is $\hat{i}+2\hat{j}-3\hat{k}$ and parallel to the vector $2\hat{i}+\hat{j}+2\hat{k}$. A plane $\mathrm{\pi }$ is passing through the points $\hat{i}+\hat{j}+\hat{k}, \hat{i}-\hat{j}-\hat{k}$ and parallel to the vector $\hat{i}-2\hat{j}$. Then the point where this plane $\mathrm{\pi }$ meets the line $L$ is

The line $\frac{x-1}{1}=\frac{y-2}{-2}=\frac{z-1}{3}$ and the plane $x+2y+z=6$ meet at 

Equation of plane which passes through the point of intersection of lines $\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}$ and $\frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ and at greatest distance from the point $\left(0,0,0\right)$ is

The angle between the line $\overrightarrow{r}=\left(2\hat{i}-\hat{j}+\hat{k}\right)+\lambda \left(-\hat{i}+\hat{j}+\hat{k}\right)$ and the plane $\overrightarrow{r}.\left(3\hat{i}+2\hat{j}-\hat{k}\right)=4$ is 

The distance of the point $\left(1,-2,3\right)$ from the plane $x-y+z-5=0$ measured parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z-1}{-6}$ is equal to