Find the distance of the point $(2, 3, -5)$ from the plane $x + 2y - 2z = 9$.

Let $P$ be the point of intersection of the line $\frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}$ and the plane $x + y + z = 2$. If the distance of the point $P$ from the plane $3x - 4y + 12z = 32$ is $q$,then $q$ and $2q$ are the roots of the equation:

$A$ plane $P$ contains the line $x+2y+3z+1=0=x-y-z-6$ and is perpendicular to the plane $-2x+y+z+8=0$. Then which of the following points lies on $P$?

The shortest distance between the line $r = 2\hat{i} - 2\hat{j} + 3\hat{k} + \lambda(\hat{i} - \hat{j} + 4\hat{k})$ and the plane $r \cdot (\hat{i} + 5\hat{j} + \hat{k}) = 5$ is

Let a unit vector $\hat{OP}$ make angles $\alpha, \beta, \gamma$ with the positive directions of the coordinate axes $OX, OY, OZ$ respectively,where $\beta \in (0, \frac{\pi}{2})$. If $\hat{OP}$ is perpendicular to the plane passing through the points $(1, 2, 3)$,$(2, 3, 4)$,and $(1, 5, 7)$,then which one of the following is true?

If the angle between the line $x = \frac{y - 1}{2} = \frac{z - 3}{\lambda}$ and the plane $x + 2y + 3z = 4$ is $\cos^{-1}\left(\sqrt{\frac{5}{14}}\right)$,then $\lambda$ equals: