The coordinates of the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane $2x - 3y - 6z = 49$ are

The mirror image of $P(2, 4, -1)$ in the plane $x - y + 2z - 2 = 0$ is $Q(a, b, c)$. Then the value of $a + b + c$ is:

The distance of the point $O(\vec{0})$ from the plane $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=5$ measured parallel to the vector $2 \hat{i}+3 \hat{j}-6 \hat{k}$ 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 $x y z=$

The distance between the line $\bar{r} = 3\hat{i} - 2\hat{j} + \hat{k} + \lambda(\hat{i} - 2\hat{j})$ and the plane $\bar{r} \cdot (2\hat{i} + \hat{j} + \hat{k}) = 4$ is

The line passing through the point $2\bar{a}+\bar{b}$ and parallel to the vector $\bar{b}-\bar{c}$ and the plane passing through the point $\bar{a}$ and parallel to the vectors $\bar{b}+\bar{c}$ and $\bar{a}+2\bar{b}-\bar{c}$ intersect at $P$. The position vector of $P$ is