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

The image of the point with position vector $(\hat{i}+3 \hat{j}+4 \hat{k})$ in the plane $r \cdot(2 \hat{i}-\hat{j}+\hat{k})+3=0$ is

If the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}$ meets the plane $x+2y+3z=15$ at the point $P$,then the distance of $P$ from the origin is

The plane $2x - y + 3z + 5 = 0$ is rotated through $90^o$ about its line of intersection with the plane $5x - 4y - 2z + 1 = 0$. The equation of the plane in its new position is:

Let $P$ be the plane passing through the intersection of the planes $\overrightarrow{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 5$ and $\overrightarrow{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 3$,and the point $(2, 1, -2)$. Let the position vectors of the points $X$ and $Y$ be $\hat{i} - 2\hat{j} + 4\hat{k}$ and $5\hat{i} - \hat{j} + 2\hat{k}$ respectively. Then the points:

The angle between the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and the plane $3x + 2y - 3z = 4$ is ......... $^o$