Let the foot of the perpendicular from the point $P (3, -2, -9)$ on the plane passing through the points $A (-1, -2, -3)$,$B (9, 3, 4)$,and $C (9, -2, 1)$ be $Q(\alpha, \beta, \gamma)$. Then the distance of $Q$ from the origin is:

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 $L_2: \frac{x-1}{0} = \frac{y-1}{1} = \frac{z-1}{-1}$.

Let $P_1$ and $P_2$ be two planes given by $P_1: 10x + 15y + 12z - 60 = 0$ and $P_2: -2x + 5y + 4z - 20 = 0$. Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on $P_1$ and $P_2$?
$(A) \frac{x-1}{0} = \frac{y-1}{0} = \frac{z-1}{5}$
$(B) \frac{x-6}{-5} = \frac{y}{2} = \frac{z}{3}$
$(C) \frac{x}{-2} = \frac{y-4}{5} = \frac{z}{4}$
$(D) \frac{x}{1} = \frac{y-4}{-2} = \frac{z}{3}$

The equation of the planes passing through the line of intersection of the planes $3x - y - 4z = 0$ and $x + 3y + 6 = 0$ whose distance from the origin is $1$,are

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

Find the equation of the plane which is perpendicular to the plane $5x + 3y + 6z + 8 = 0$ and which contains the line of intersection of the planes $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 = 0$.