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

Assertion $(A)$: The equation of the plane passing through the point $(4, 4, 4)$ and the intersection of the planes $x + y + z = 6$ and $2x + 3y + 4z = 0$ is $29x + 23y + 17z = 276$.
Reason $(R)$: The equation of the plane passing through the line of intersection of planes $P_1 = 0$ and $P_2 = 0$ is $P_1 + \lambda P_2 = 0, \lambda \in \mathbb{R}$.

Let $A(2,5,7)$ be the image of the point $B(1,-2,3)$ with respect to a plane $\pi$. Let $C$ be the point where $AB$ meets the plane $\pi$. Let $D=(2,1,6)$. Then the direction cosines of $CD$ are

The equation of the plane passing through the intersection of the planes $2x - 5y + z = 3$ and $x + y + 4z = 5$ and parallel to the plane $x + 3y + 6z = 1$ is $x + 3y + 6z = k$,where $k$ is:

If the line of intersection of the planes $ax + by = 3$ and $ax + by + cz = 0$ $(a > 0)$ makes an angle $30^{\circ}$ with the plane $y - z + 2 = 0$,then the direction cosines of the line are:

$\overrightarrow{AB} = 3\hat{i} - \hat{j} + \hat{k}$ and $\overrightarrow{CD} = -3\hat{i} + 2\hat{j} + 4\hat{k}$ are two vectors. The position vectors of the points $A$ and $C$ are $6\hat{i} + 7\hat{j} + 4\hat{k}$ and $-9\hat{j} + 2\hat{k}$ respectively. Find the position vectors of a point $P$ on the line $AB$ and a point $Q$ on the line $CD$ such that $\overrightarrow{PQ}$ is perpendicular to both $\overrightarrow{AB}$ and $\overrightarrow{CD}$.