The line of intersection of the planes $x + 2y = 0$ and $y - 3z + 3 = 0$ is

Let a unit vector $\hat{OP}$ make angles $\alpha, \beta, \gamma$ with the positive directions of the coordinate axes $OX, OY, OZ$ respectively,where $\beta \in (0, \frac{\pi}{2})$. If $\hat{OP}$ is perpendicular to the plane passing through the points $(1, 2, 3)$,$(2, 3, 4)$,and $(1, 5, 7)$,then which one of the following is true?

The direction ratios of the normal to the plane passing through the points $(1, 2, -3)$,$(-1, -2, 1)$ and parallel to $\frac{x-2}{2}=\frac{y+1}{3}=\frac{z}{4}$ are:

Let $L_1: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}$ and $L_2: \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}$ be two lines. Let $L_3$ be a line passing through the point $(\alpha, \beta, \gamma)$ and be perpendicular to both $L_1$ and $L_2$. If $L_3$ intersects $L_1$,then $|5\alpha-11\beta-8\gamma|$ equals :

The direction cosines of the line formed by the intersection of the planes $x - y + 2z = 5$ and $3x + y + z = 6$ are:

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