If the plane $4x + 4y - kz = 0$ is the equation of the plane passing through the origin and containing the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$,find the value of $k$.

Equation of the plane containing the line $x + 2y + 3z - 5 = 0 = 3x + 2y + z - 5$ which is parallel to the line $x - 1 = 2 - y = z - 3$ is:

If the image of the point $P(1, -2, 3)$ in the plane $2x + 3y - 4z + 22 = 0$ measured parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$ is $Q$,then $PQ$ is equal to:

The equation of the line passing through $(-4, 1, 3)$,parallel to the plane $x + 2y - z - 5 = 0$ and intersecting the line $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z - 2}{-1}$ is

Let the equation of the plane,that passes through the point $(1,4,-3)$ and contains the line of intersection of the planes $3x-2y+4z-7=0$ and $x+5y-2z+9=0$,be $\alpha x+\beta y+\gamma z+3=0$. Then $\alpha+\beta+\gamma$ is equal to:

The coordinates of the foot of the perpendicular from the point $(1, -2, 1)$ on the plane containing the lines $\frac{x + 1}{6} = \frac{y - 1}{7} = \frac{z - 3}{8}$ and $\frac{x - 1}{3} = \frac{y - 2}{5} = \frac{z - 3}{7}$ is