If lines $\frac{x-3}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-1}{1}=\frac{6-z}{5}$ are perpendicular to each other,then $k=$ $\qquad$ .

The foot of the perpendicular from the point $(1, 2, 3)$ on the line $\vec{r} = (6 \hat{i} + 7 \hat{j} + 7 \hat{k}) + \lambda(3 \hat{i} + 2 \hat{j} - 2 \hat{k})$ has the coordinates:

Let the point $A$ be the foot of the perpendicular drawn from the point $P(a, b, 0)$ to the line $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-\alpha}{3}$. If the midpoint of the line segment $PA$ is $(0, \frac{3}{4}, -\frac{1}{4})$, then the value of $a^2+b^2+\alpha^2$ is equal to:

Find the perpendicular distance from the point $P(4, -5, 3)$ to the line $\vec{r} = (5, -2, 6) + k(3, -4, 5)$,where $k \in \mathbb{R}$.

The coordinates of the foot of the perpendicular drawn from the point $2 \hat{i} - \hat{j} + 5 \hat{k}$ to the line $\vec{r} = (11 \hat{i} - 2 \hat{j} - 8 \hat{k}) + \lambda(10 \hat{i} - 4 \hat{j} - 11 \hat{k})$ are

The angle between the lines $3x = 2y = -z$ and $-x = 6y = -4z$ is