Two lines $L_1: x=5, \frac{y}{3-\alpha}=\frac{z}{-2}$ and $L_2: x=\alpha, \frac{y}{-1}=\frac{z}{2-\alpha}$ are coplanar. Then $\alpha$ can take value$(s)$.

What are the direction ratios of the line represented by the equations $x = ay + b$ and $z = cy + d$?

Find the values of $p$ so that the lines $\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}$ and $\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles.

The length of the perpendicular drawn from the point $P(5, 4, -1)$ to the line $\frac{x - 1}{2} = \frac{y}{9} = \frac{z}{5}$ is:

The equation of the line equally inclined with the positive coordinate axes and passing through $(-3, 2, -5)$ is

The line $L$ passes through the point $(1, 2, 3)$. The distance of any point on the line $L$ from the line $\vec{r} = (-1, 3, 4) + \lambda(3, -2, 1)$ is constant. Then the line $L$ does not pass through the point: