The line segment joining the points $A(2, 3, 4)$ and $B(-3, 5, -4)$ intersects the $yz$-plane at the point:

If the direction ratios of two lines are given by $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0$,then the angle between the lines is

Let a line $L$ be perpendicular to both the lines $L_1: \frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}$ and $L_2: \frac{x-2}{1} = \frac{y-4}{4} = \frac{z-6}{7}$. If $\theta$ is the acute angle between the lines $L$ and $L_3: \frac{x-7}{2} = \frac{y-7}{1} = \frac{z}{2}$, then $\tan \theta$ is equal to:

$A$ triangle $ABC$ is formed by vertices $A(1, -1, 0)$,$B(3, 5, 3)$,and $C(-11, -5, 6)$. The equation of the internal angle bisector of $\angle A$ is:

Show that the points $A(2, 3, -4)$,$B(1, -2, 3)$,and $C(3, 8, -11)$ are collinear.

The angle between the two lines $\frac{x-2}{2} = \frac{2-y}{3} = \frac{z-1}{2}$ and $\frac{x-1}{2} = \frac{y+1}{1} = \frac{z-3}{-3}$ is . . . . . . .