If the lines $\frac{1-x}{3} = \frac{3y-6}{k} = \frac{3-z}{-2}$ and $\frac{1-x}{2k} = \frac{y-5}{3} = \frac{6-z}{5}$ are perpendicular to each other, then $k$ is equal to

A line makes the angle $\theta$ with each of the $x$ and $z$ axes. If the angle $\beta$ which it makes with $y$-axis is such that $\sin^2\beta = 3\sin^2\theta$, then the value of $\cos^2\theta$ is

The image of the point (1, 6, 3) with respect to the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ is

The distance of the point $(2, 3, -5)$ from the plane $x + 2y - 2z = 9$ is :

The points of trisection of the segment joining the points (1, 0, 2) and (1, 3, 2) are:

A. $(1, \frac{3}{2}, \frac{4}{3})$

B. (1, 1, 2)

C. (1, 2, 2)

D. $(1, \frac{3}{2}, 2)$

Choose the correct answer from the options given below:

The angle between the lines $\dfrac{3-x}{-7} = \dfrac{y+5}{-5} = \dfrac{z}{1}$ and $\dfrac{x}{1} = \dfrac{y}{2} = \dfrac{z}{3}$ is:

The angle between the two planes $x + y - z = 3$ and $3x + 2y + z = 5$ is :