The shortest distance between the skew lines $r=(6 \hat{i}+2 \hat{j}+2 \hat{k})+t(\hat{i}-2 \hat{j}+2 \hat{k})$ and $r=(-4 \hat{i}-\hat{k})+s(3 \hat{i}-2 \hat{j}-2 \hat{k})$ is

The two lines $L_1: \vec{r}=(\hat{i}+5 \hat{j}+5 \hat{k})+t(4 \hat{i}-4 \hat{j}+5 \hat{k})$ and $L_2: \vec{r}=(2 \hat{i}+4 \hat{j}+5 \hat{k})+s(8 \hat{i}-3 \hat{j}+\hat{k})$ are such that

If the lines $\frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2}$ and $\frac{x - 1}{3k} = \frac{y - 5}{1} = \frac{z - 6}{-5}$ are at right angles,then $k =$

The line $l_1$ passes through the point $(2, 6, 2)$ and is perpendicular to the plane $2x + y - 2z = 10$. Then the shortest distance between the line $l_1$ and the line $\frac{x + 1}{2} = \frac{y + 4}{-3} = \frac{z}{2}$ is :

The equation of the line passing through $(3, -1, 2)$ and perpendicular to the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(2 \hat{i} - 2 \hat{j} + \hat{k})$ and $\vec{r} = (2 \hat{i} + \hat{j} - 3 \hat{k}) + \mu(\hat{i} - 2 \hat{j} + 2 \hat{k})$ is

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.