Point $A$ lies at a distance of $6$ units from the point $(1, 0, 1)$ on the line $\frac{x - 1}{2} = \frac{y}{2} = \frac{z - 1}{1}$ in the negative $z$-direction. Find the coordinates of $A$.

The lines whose direction cosines satisfy the equations $al + bm + cn = 0$ and $fmn + gnl + hlm = 0$ will be perpendicular if...

Let a straight line $L$ pass through the point $P(2, -1, 3)$ and be perpendicular to the lines $\frac{x-1}{2} = \frac{y+1}{1} = \frac{z-3}{-2}$ and $\frac{x-3}{1} = \frac{y-2}{3} = \frac{z+2}{4}$. If the line $L$ intersects the $yz$-plane at the point $Q$,then the distance between the points $P$ and $Q$ is:

For what value of $\lambda$ are the lines $\frac{x - 1}{1} = \frac{y - 2}{\lambda} = \frac{z + 1}{-1}$ and $\frac{x + 1}{-\lambda} = \frac{y + 1}{2} = \frac{z - 2}{1}$ perpendicular to each other?

The vector equation of the line whose Cartesian equations are $y=2$ and $4x-3z+5=0$ is

If the lines $2x = ky = -z$ and $6x = -y = -4z$ are perpendicular to each other, then the value of $k$ is: