$A$ ray of light passing through the point $A(1, 2, 3)$ strikes the plane $x+y+z=12$ at $B$ and on reflection passes through $C(3, 5, 9)$,then $OB$ is equal to

$A$ 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 = $

If the lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar,then the equation of the plane containing these lines is:

The line $\ell$ passes through the point $P(2, 1, 1)$ and is parallel to the plane $x + y + 2z = 18$. If line $\ell$ intersects the line $L_2: \frac{x + 2}{3} = \frac{y + 1}{-1} = \frac{z - 2}{1}$, then the equation of the line $\ell$ is:

The equation of the plane passing through the line of intersection of the planes $x + y + z = 5$ and $2x + 3y + 4z + 5 = 0$ and perpendicular to the plane $x + y + z = 5$ is

The distance of the point $(1, 3, -7)$ from the plane passing through the point $(1, -1, -1)$ and having a normal perpendicular to both the lines $\frac{x - 1}{1} = \frac{y + 2}{-2} = \frac{z - 4}{3}$ and $\frac{x - 2}{2} = \frac{y + 1}{-1} = \frac{z + 7}{-1}$ is . . . .