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?

Let the foot of the perpendicular from the point $(\lambda, 2, 3)$ on the line $\frac{x-4}{1} = \frac{y-9}{2} = \frac{z-5}{1}$ be the point $(1, \mu, 2)$. Then the distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}$ and $\frac{x-\lambda}{2} = \frac{y-\mu}{3} = \frac{z+5}{6}$ is equal to:

The lines $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}$ and $\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{2}$

The angle between the lines $\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and $\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}$ is . . . . . . .

The line $L_1$ is parallel to the vector $\vec{a} = -3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and passes through the point $(7, 6, 2)$,and the line $L_2$ is parallel to the vector $\vec{b} = 2 \hat{i} + \hat{j} + 3 \hat{k}$ and passes through the point $(5, 3, 4)$. The shortest distance between the lines $L_1$ and $L_2$ is:

If the lines $\vec{r} = \hat{i} + \hat{j} - \hat{k} + \lambda(q\hat{i} - 2\hat{j} + \hat{k})$ and $\vec{r} = p\hat{i} - 3\hat{j} + 2\hat{k} + \mu(\hat{i} - 2\hat{j} + 2\hat{k})$ intersect each other and $q\hat{i} - 2\hat{j} + \hat{k}$ is collinear to $4\hat{i} - 4\hat{j} + 2\hat{k}$, then the values of $p$ and $q$ are