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

$\pi_1$ is a plane passing through the point $(1, 2, 3)$ and perpendicular to the planes $x+2y+3z-6=0$ and $x+2y+2z-5=0$. If $(-1, 2, -3)$ is the foot of the perpendicular drawn from the point $(1, 3, 2)$ onto a plane $\pi_2$,then the angle between the planes $\pi_1$ and $\pi_2$ is

Find the distance of a point $(2, 5, -3)$ from the plane $\vec{r} \cdot (6 \hat{i} - 3 \hat{j} + 2 \hat{k}) = 4$. (in $/7$)

$A$ plane meets the coordinate axes at $P, Q, R$ respectively. If the centroid of $\triangle P Q R$ is $\left(1, \frac{1}{2}, \frac{1}{3}\right)$,then the equation of the plane 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 :

The value of $a$ for which the three given planes
$P_1 : (a + 1)x - y - z = a$
$P_2 : x - (a + 1)y - z = a$
$P_3 : x - y - (a + 1)z = a$
have no point in common,is