If $4x + 4y - kz = 0$ is the equation of the plane through the origin that contains the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4},$ then $k =$

The image of the point $(3, 2, 1)$ in the plane $2x - y + 3z = 7$ is

Equation of the plane which passes through the point of intersection of lines $\frac{x - 1}{3} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ and has the largest distance from the origin is

Find the angle between the plane $\bar{r} \cdot (1, 2, 1) = 1$ and the line $\frac{x}{2} = \frac{y}{1} = \frac{z}{-1}$.

The angle between the line $\frac{x - 2}{a} = \frac{y - 2}{b} = \frac{z - 2}{c}$ and the plane $ax + by + cz + 6 = 0$ is ......... $^o$

The equation of the straight line passing through $(1, 2, 3)$ and perpendicular to the plane $x + 2y - 5z + 9 = 0$ is