Find the distance between the line $\vec{r} = (2\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + 4\hat{k})$ and the plane $\vec{r} \cdot (\hat{i} + 5\hat{j} + \hat{k}) = 5$.

Let the line $L$ be the projection of the line $\frac{x-1}{2}=\frac{y-3}{1}=\frac{z-4}{2}$ in the plane $x-2y-z=3$. If $d$ is the distance of the point $(0,0,6)$ from $L$,then $d^2$ is equal to .... .

Let $P$ be the plane,which contains the line of intersection of the planes $x + y + z - 6 = 0$ and $2x + 3y + z + 5 = 0$ and it is perpendicular to the $xy$-plane. Then the distance of the point $(0, 0, 256)$ from $P$ is equal to

The acute angle $\theta$ between the $xy$-plane and the plane passing through the point $(1, 2, 4)$ and parallel to the vectors with direction ratios $3, 2, -1$ and $1, -2, -2$ is

The value of $k$,such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ lies on the plane $2x-4y+z=7$,is

$\vec{a}, \vec{b}, \vec{c}$ 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 $x y z=$