Let $Q$ be the foot of the perpendicular from the origin to the plane $4x - 3y + z + 13 = 0$ and $R$ be a point $(-1, 1, -6)$ on the plane. Then the length $QR$ is

If $(2, -1, 3)$ is the foot of the perpendicular drawn from the origin $(0, 0, 0)$ to a plane,then the equation of that plane is:

$P$ is a fixed point $(a, a, a)$ on a line through the origin equally inclined to the axes. Then,any plane through $P$ perpendicular to $OP$ makes intercepts on the axes,the sum of whose reciprocals is equal to:

The distance of a point $\vec{a}$ from the plane $\vec{r} \cdot \vec{m} = q$ is given by $\frac{|\vec{a} \cdot \vec{m} - q|}{|\vec{m}|}$. If the distance of the point $\hat{i} + 2\hat{j} + 3\hat{k}$ from the plane $\vec{r} \cdot (2\hat{i} + 6\hat{j} - 9\hat{k}) = -1$ is $p$ and the distance of the origin from this plane is $q$,then $p - q =$

The equation of the plane passing through the point $(2, 3, 4)$ and parallel to the plane $x + 2y + 4z = 5$ is:

If $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are three non-coplanar vectors, then the vector equation $\overrightarrow{r}=(1-p-q) \overrightarrow{a}+p \overrightarrow{b}+q \overrightarrow{c}$ represents a :