If the shortest distance between the line joining the points $(1, 2, 3)$ and $(2, 3, 4)$,and the line $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-2}{0}$ is $\alpha$,then $28 \alpha^2$ is equal to $........$.

If the lines $\frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2}$ and $\frac{x - 1}{3k} = \frac{y - 5}{1} = \frac{z - 6}{-5}$ are perpendicular to each other,then $k = \dots$

If $(2\alpha + 1, \alpha^2 - 3\alpha, \frac{\alpha - 1}{2})$ is the image of $(\alpha, 2\alpha, 1)$ in the line $\frac{x-2}{3} = \frac{y-1}{2} = \frac{z}{1}$, then the possible value$(s)$ of $\alpha$ is (are)

Find the foot of the perpendicular drawn from the point $A(1, 0, 3)$ to the line joining the points $B(4, 7, 1)$ and $C(3, 5, 3)$.

The distance of the point $P(4, 6, -2)$ from the line passing through the point $(-3, 2, 3)$ and parallel to a line with direction ratios $3, 3, -1$ is equal to:

Statement-$1$: The shortest distance between the skew lines $\frac{x+3}{-4} = \frac{y-6}{3} = \frac{z}{2}$ and $\frac{x+3}{-4} = \frac{y}{1} = \frac{z-7}{1}$ is $9$.
Statement-$2$: Two lines are skew lines if there exists no plane passing through them.