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

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} + \hat{k}$ and $\overrightarrow{b} = 2\hat{i} + 7\hat{j} + 3\hat{k}$. Let $L_1: \overrightarrow{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda \overrightarrow{a}, \lambda \in R$ and $L_2: \overrightarrow{r} = (\hat{j} + \hat{k}) + \mu \overrightarrow{b}, \mu \in R$ be two lines. If the line $L_3$ passes through the point of intersection of $L_1$ and $L_2$,and is parallel to $\overrightarrow{a} + \overrightarrow{b}$,then $L_3$ passes through the point:

The line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{-1}$ intersects the curve $xy = c^2, z = 0$ if $c$ is equal to

Find the distance of the point $(-1, -2, -1)$ from the plane passing through the point $(1, 1, 1)$ and perpendicular to both lines $L_1: \frac{x-1}{1} = \frac{y-1}{0} = \frac{z-1}{-1}$ and $L_2: \frac{x-1}{0} = \frac{y-1}{1} = \frac{z-1}{-1}$.

If the line joining the points $\overline{i} + 2\overline{j}$ and $\overline{j} - 2\overline{k}$ intersects the plane passing through the points $2\overline{i} - \overline{j}$,$2\overline{j} + 3\overline{k}$,and $\overline{k} - 2\overline{i}$ at $\overline{r}$,then $\overline{r} \cdot (\overline{i} + \overline{j} + \overline{k}) = $

$A$ line with positive direction cosines passes through the point $P(2,1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $2x+y+z=9$ at point $Q$. The length of the line segment $PQ$ equals $\qquad$ units.