Let the shortest distance between the lines $\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}$ be $3 \sqrt{30}$. Then the positive value of $5 \alpha+\beta$ is

If lines $\frac{2x-4}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}$ and $\frac{x-1}{1}=\frac{3y-1}{\lambda}=\frac{z-2}{1}$ are perpendicular to each other,then $\lambda = \ldots$.

The equation of a line in Cartesian form passing through $(0, 0, 0)$ and $(4, 3, c)$ and parallel to $a \times b$ where $a = 2i + j + 2k$ and $b = 3i - 4j$ is given by the direction vector $a \times b$. Find the value of $c$ and the equation of the line.

The Cartesian equations of the line passing through the point $A(0, 1, 1)$ and parallel to the $X$-axis are:

$A$ line $L_1$ passes through the point with position vector $3 \hat{i}$ and is parallel to the vector $-\hat{i}+\hat{j}+\hat{k}$. Another line $L_2$ passes through the point with position vector $\hat{i}+\hat{j}$ and is parallel to the vector $\hat{i}+\hat{k}$. Find the position vector of the point of intersection of lines $L_1$ and $L_2$.

The lines $\bar{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3 \hat{i} - \hat{j})$ and $\bar{r} = (4 \hat{i} - \hat{k}) + \mu(2 \hat{i} + 3 \hat{k})$ are