The shortest distance between the lines $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$ and $\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$ is

If the foot of the perpendicular drawn from the point $A(1, 0, 3)$ on a line passing through $B(\alpha, 7, 1)$ is $P\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)$,then $\alpha$ is equal to:

Let the line $L$ intersect the lines $x-2=-y=z-1$ and $2(x+1)=2(y-1)=z+1$,and be parallel to the line $\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$. Then which of the following points lies on $L$?

If the shortest distance between the lines $\frac{x-\lambda}{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ and $\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}$ is $1$,then the sum of all possible values of $\lambda$ is:

The equation of the line passing through the point $(1, 2, 3)$ and perpendicular to the lines $\frac{x-2}{3} = \frac{y-1}{2} = \frac{z+1}{-2}$ and $\frac{x}{2} = \frac{y}{-3} = \frac{z}{1}$ is

If the foot of the perpendicular drawn from $(1, 9, 7)$ to the line passing through the point $(3, 2, 1)$ and parallel to the planes $x+2y+z=0$ and $3y-z=3$ is $(\alpha, \beta, \gamma)$,then $\alpha+\beta+\gamma$ is equal to