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

Let $P$ be a plane $lx + my + nz = 0$ containing the line $\frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3}$. If plane $P$ divides the line segment $AB$ joining points $A(-3, -6, 1)$ and $B(2, 4, -3)$ in the ratio $k : 1$,then the value of $k$ is equal to

Let $Q$ be the mirror image of the point $P(1, 0, 1)$ with respect to the plane $S: x + y + z = 5$. If a line $L$ passing through $(1, -1, -1)$,parallel to the line $PQ$,meets the plane $S$ at $R$,then $QR^{2}$ is equal to

The equation of the line passing through $(1, 2, 3)$ and parallel to the planes $x - y + 2z = 5$ and $3x + y + z = 6$ is:

Let $P$ be the image of the point $(3, 1, 7)$ with respect to the plane $x-y+z=3$. Then the equation of the plane passing through $P$ and containing the straight line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$ is

If the perpendicular distance of the plane passing through the point $Q(1, 0, -1)$ and containing the line $\vec{r} = (\hat{i} - 3\hat{j} + \hat{k}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k})$ from the origin is $\frac{p}{\sqrt{53}}$, then $p = $