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 angle between the pair of lines $\vec{r} = -3\hat{i} + \hat{j} + 3\hat{k} + \lambda(3\hat{i} + 5\hat{j} + 4\hat{k})$ and $\vec{r} = -\hat{i} + 4\hat{j} + 5\hat{k} + \mu(\hat{i} + \hat{j} + 2\hat{k})$ is . . . . . . .

Let $(\alpha, \beta, \gamma)$ be the coordinates of the foot of the perpendicular drawn from the point $(5, 4, 2)$ on the line $\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$. Then the length of the projection of the vector $\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$ on the vector $6\hat{i} + 2\hat{j} + 3\hat{k}$ is:

If two lines $L_1$ and $L_2$ in space are defined by $L_1 = \{ x = \sqrt{\lambda} y + (\sqrt{\lambda} - 1), z = (\sqrt{\lambda} - 1)y + \sqrt{\lambda} \}$ and $L_2 = \{ x = \sqrt{\mu} y + (1 - \sqrt{\mu}), z = (1 - \sqrt{\mu})y + \sqrt{\mu} \}$,then $L_1$ is perpendicular to $L_2$ for all non-negative reals $\lambda$ and $\mu$ such that:

Find the vector and the cartesian equations of the lines that passes through the origin and $(5, -2, 3).$

Show that the points $A(2, 3, -4)$,$B(1, -2, 3)$,and $C(3, 8, -11)$ are collinear.