The point on the plane $2x - 2y + 4z + 5 = 0$ that is nearest to $\left(1, \frac{3}{2}, 2\right)$ is

For which values of $a$ do the two points $(1, a, 1)$ and $(-3, 0, a)$ lie on opposite sides of the plane $3x + 4y - 12z + 13 = 0$?

$A$ variable plane is at a constant distance $h$ from the origin and meets the coordinate axes in $A, B, C$. The locus of the centroid of $\triangle ABC$ is

The length of the perpendicular from the origin to the plane passing through three non-collinear points $a, b, c$ is

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 $p - q =$

Assertion: The points $(2, 1, 5)$ and $(3, 4, 3)$ lie on opposite sides of the plane $2x + 2y - 2z - 1 = 0$.
Reason: The algebraic perpendicular distances from the given points to the plane have opposite signs.