Find the vector equation of the line passing through the point $(1, 2, -4)$ and perpendicular to the two lines: $\frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7}$ and $\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}$.

The sum of all values of $ \alpha $, for which the shortest distance between the lines $ \frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha} $ and $ \frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2\alpha} $ is $ \sqrt{2} $, is

The line joining points $(3, 5, -7)$ and $(-2, 1, 8)$ meets the $yz$-plane at which point?

Let $L_1$ and $L_2$ denote the lines $\overrightarrow{r} = \hat{i} + \lambda(-\hat{i} + 2\hat{j} + 2\hat{k}), \lambda \in R$ and $\overrightarrow{r} = \mu(2\hat{i} - \hat{j} + 2\hat{k}), \mu \in R$ respectively. If $L_3$ is a line which is perpendicular to both $L_1$ and $L_2$ and intersects both of them,then which of the following options describe$(s)$ $L_3$?
$(1) \overrightarrow{r} = \frac{1}{3}(2\hat{i} + \hat{k}) + t(2\hat{i} + 2\hat{j} - \hat{k}), t \in R$
$(2) \overrightarrow{r} = \frac{2}{9}(2\hat{i} - \hat{j} + 2\hat{k}) + t(2\hat{i} + 2\hat{j} - \hat{k}), t \in R$
$(3) \overrightarrow{r} = t(2\hat{i} + 2\hat{j} - \hat{k}), t \in R$
$(4) \overrightarrow{r} = \frac{2}{9}(4\hat{i} + \hat{j} + \hat{k}) + t(2\hat{i} + 2\hat{j} - \hat{k}), t \in R$

The coordinates of the point where the line passing through $P(3, 4, 1)$ and $Q(5, 1, 6)$ crosses the $xy$-plane are:

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