If the distance between the plane $ax - 2y + z = k$ and the plane containing the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}$ is $\sqrt{6}$, then $|k|$ is

If the point of intersection of the lines $r = \hat{i} - 6\hat{j} + (p \sec \alpha) \hat{k} + t(\hat{i} + 2\hat{j} + \hat{k})$ and $r = 4\hat{j} + \hat{k} + \lambda(2\hat{i} + (p \tan \alpha) \hat{j} + 2\hat{k})$ is $8\hat{i} + 8\hat{j} + 9\hat{k}$, (where $0 < \alpha < \frac{\pi}{2}$), then $p =$

In what ratio does the plane $ax + by + cz + d = 0$ divide the line segment joining the points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$?

$\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors. If the position vector of the point of intersection of the line $\vec{r}=\vec{a}+2 \vec{b}+p(\vec{a}-2 \vec{c})$ and the plane $\vec{r}=3 \vec{a}-q(\vec{c}-\vec{b})+k(\vec{a}-\vec{b}+\vec{c})$ is $\vec{r}=x \vec{a}+y \vec{b}+z \vec{c}$, then $x y z=$

If the line $r = a + t b$ is parallel to the plane $r = c + l d + m e$, then

Three lines are given by $\overrightarrow{r} = \lambda \hat{i}, \lambda \in R$,$\overrightarrow{r} = \mu(\hat{i} + \hat{j}), \mu \in R$ and $\overrightarrow{r} = v(\hat{i} + \hat{j} + \hat{k}), v \in R$. Let the lines cut the plane $x + y + z = 1$ at the points $A, B$ and $C$ respectively. If the area of the triangle $ABC$ is $\Delta$,then the value of $(6 \Delta)^2$ equals.