The distance of the line $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$ from the point $(1,4,0)$ along the line $\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}$ is:

The image of the point $(5, 2, 6)$ with respect to the plane $x + y + z = 9$ is

Find the equation of the plane passing through the intersection of the planes $P_1$ and $P_2$ and parallel to the line $L$,where:
$P_1 : 3x + 2y + 5z + 1 = 0$
$P_2 : x + y + z + 2 = 0$
$L : \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$

The equation of a plane containing the line of intersection of the planes $2x - y - 4 = 0$ and $y + 2z - 4 = 0$ and passing through the point $(1, 1, 0)$ is

$A$ line $L$ passes through the points $(1, 2, -3)$ and $(3, 3, -1)$ and a plane $\pi$ passes through the points $(2, 1, -2), (-2, -3, 6)$,and $(0, 2, -1)$. If $\theta$ is the angle between the line $L$ and plane $\pi$,then $27 \cos^2 \theta = $

The value of $k$,such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ lies on the plane $2x-4y+z=7$,is