The value of $\lambda$ so that the lines $\frac{1-x}{3} = \frac{7y-14}{2\lambda} = \frac{z-3}{2}$ and $\frac{7-7x}{3\lambda} = \frac{y-5}{1} = \frac{6-z}{5}$ are at right angle, is:

Match List-I with List-II

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The co-ordinates of the point at which the line $\frac{x-3}{3} = \frac{y+1}{2} = \frac{z-4}{-2}$ crosses x-y plane, are

The vector equation of line passing through (2, -1, 3) and perpendicular to the lines

$\frac{x-2}{3} = \frac{y-1}{1} = \frac{z+2}{2}$ and $\frac{x+3}{-4} = \frac{y-5}{-3} = \frac{z+1}{2}$ is

(Here $\lambda$ is a parameter)

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

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

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