If 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 angles,then $\lambda=$

The angle between the lines $3x = 2y = -z$ and $-x = 6y = -4z$ is

If two lines $L_1$ and $L_2$ in space are defined by $L_1 = \{ x = \sqrt{\lambda} y + (\sqrt{\lambda} - 1), z = (\sqrt{\lambda} - 1)y + \sqrt{\lambda} \}$ and $L_2 = \{ x = \sqrt{\mu} y + (1 - \sqrt{\mu}), z = (1 - \sqrt{\mu})y + \sqrt{\mu} \}$,then $L_1$ is perpendicular to $L_2$ for all non-negative reals $\lambda$ and $\mu$ such that:

One vertex of a rectangular parallelepiped is at the origin $O$ and the lengths of its edges along $x, y$ and $z$ axes are $3, 4$ and $5$ units respectively. Let $P$ be the vertex $(3, 4, 5)$. Then the shortest distance between the diagonal $OP$ and an edge parallel to the $z$-axis,not passing through $O$ or $P$,is:

If the length of the perpendicular from the point $P(\beta, 0, \beta) \, (\beta \neq 0)$ to the line $\frac{x}{1} = \frac{y - 1}{0} = \frac{z + 1}{-1}$ is $\sqrt{\frac{3}{2}}$,then $\beta$ is equal to

The coordinates of the foot of the perpendicular drawn from the origin to the line joining the points $(-9, 4, 5)$ and $(10, 0, -1)$ are