The shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ is

If the angle between the lines,$\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{5 - x}{- 2} = \frac{7y - 14}{p} = \frac{z - 3}{4}$ is $\cos^{-1} \left( \frac{2}{3} \right)$,then $p$ is equal to

Let $(\alpha, \beta, \gamma)$ be the image of the point $A(8, 5, 7)$ in the line $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{5}$. Then $\alpha + \beta + \gamma$ is equal to

The square of the distance of the image of the point $A(6, 1, 5)$ in the line $\frac{x-1}{3} = \frac{y}{2} = \frac{z-2}{4}$ from the origin 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?

Line $L_1$ passes through the point $(1, 2, 3)$ and is parallel to the $z$-axis. Line $L_2$ passes through the point $(\lambda, 5, 6)$ and is parallel to the $y$-axis. Let for $\lambda = \lambda_1, \lambda_2$ with $\lambda_2 < \lambda_1$,the shortest distance between the two lines be $3$. Then the square of the distance of the point $(\lambda_1, \lambda_2, 7)$ from the line $L_1$ is: