The graph of the equation $y^2 + z^2 = 0$ in three-dimensional space is

Find the angle between the pair of lines $\frac{x+3}{3}=\frac{y-1}{5}=\frac{z+3}{4}$ and $\frac{x+1}{1}=\frac{y-4}{1}=\frac{z-5}{2}$.

If the line passing through the points $(-5, 1, 3)$ and $(1, 2, 0)$ is perpendicular to the line passing through the points $(x, 2, 1)$ and $(0, -4, 6)$,then $x = \dots$

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

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

Let the foot of the perpendicular from the point $(\lambda, 2, 3)$ on the line $\frac{x-4}{1} = \frac{y-9}{2} = \frac{z-5}{1}$ be the point $(1, \mu, 2)$. Then the distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}$ and $\frac{x-\lambda}{2} = \frac{y-\mu}{3} = \frac{z+5}{6}$ is equal to: