Interaction between a Line and a Plane

Find the distance of the point  $\left(-1, -5, -10\right)$, from the point of intersection of the line $\overrightarrow{r}=\left(2\widehat{i}-\widehat{j}+2\widehat{k}\right)+\lambda \left(3\widehat{i}+4\widehat{j}+2\widehat{k}\right)$ and the plane $\overrightarrow{r}\cdot \left(\widehat{i}-\widehat{j}+\widehat{k}\right)=5$.

The coordinate of the point where the line $\frac{x+1}{2}=\frac{y+2}{3}=\frac{z+3}{4}$ meets the plane $x+y+4z=6$ is

If the lines  $\frac{x-1}{-3}=\frac{y-2}{-2k}=\frac{z-3}{2}$  and  $\frac{x-1}{k}=\frac{y-2}{1}=\frac{z-3}{5}$ are perpendicular, what would be the value of $k$ and the equation of plane containing these lines?

The coordinates of the point where the line  $\frac{x+1}{2}=\frac{y+2}{3}=\frac{z+3}{4}$  meets the plane $x+y+4z=6$  would be:

The perpendicular distance from the point $\left(3,1,1\right)$ on the plane passing through the point $\left(1,2,3\right)$ and containing the line, $\overrightarrow{r}=\hat{i}+\hat{j}+\lambda \left(2\hat{i}+\hat{j}+4\hat{k}\right)$ is

If the line $\frac{x-1}{4}=\frac{y+3}{-2}=\frac{z+5}{1}$ lies in the plane $2x+ly+mz=16,$ then $l^2+m^2$ is equal to

If planes $x-cy-bz=0, cx-y+az=0$ and $bx+ay-z=0$ pass through a straight line then $a^2+b^2+c^2=$

The length of the projection of the line segment joining the points $(5,-1, 4)$ and $(4,-1, 3)$ on the plane, $x+y+z=7$ is

The distance between the point $\left(-1,-5,-10\right)$ and the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ with the plane $x-y+z=5$ is $13t,$ then $t$ equals to

The measure of the angle between the line $\overrightarrow{r}=\left(2,-3,1\right)+k\left(2,2,1\right) ;k\in R$ and the plane $2x-2y+z+7=0$ is $\dots \dots \dots ..$

The perpendicular distance from the point of intersection of the line $\frac{x+1}{2}=\frac{y+2}{3}=\frac{z}{-1}$ and plane $2x-y+z=0$ to the $Z-$axis is $\dots \dots \dots .$

Let $N\left(p,q,r\right)$ represents the point where the line through $\left(3,4,1\right)$ and $\left(5,1,0\right)$ intersects $yz$-plane, then $q-2r=$

If $\beta$ be the angle between the line $\overrightarrow{r}=\left(-\hat{i}+\hat{j}+2\hat{k}\right)+\lambda \left(\hat{i}+2\hat{j}+2\hat{k}\right)$ and the plane $\overrightarrow{r}·\left(2\hat{i}-\hat{j}+a\hat{k}\right)+4=0$ such that $\sin \beta =\frac{1}{3}$, then the value of $a$ can be

If the plane $2x – 3y – 6z = 13$ makes an angle ${\sin }^{-1}\left(\lambda \right)$ with the $x -$ axis, then the value of $\lambda$ is 

The angle between the line $\overrightarrow{r}=(\hat{i}+2\hat{ȷ}-\hat{k})+\lambda (\hat{i}-\hat{j}+\hat{k})$ and the plane $\overrightarrow{r}·(2\hat{i}-\hat{ȷ}+\hat{k})=5$ is 

The angle between the plane $\overrightarrow{r}.\left(\hat{i}-2\hat{j}+3\hat{k}\right)=5$ and the line $\overrightarrow{r}=\left(\hat{i}+\hat{j}-\hat{k}\right)+\lambda \left(\hat{i}-\hat{j}+\hat{k}\right)$ is 

The angle between the line $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z+2}{4}$ and the plane $2x+y-3z+4=0$ 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

If the line $\frac{x-3}{2}=\frac{y-4}{3}=\frac{z-5}{4}$ lies in the plane $4x+4y-kz-d=0$, then find the value of $\left(k+d\right)$.