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 $/7$)

In the following case,determine the direction cosines of the normal to the plane and the distance from the origin: $x+y+z=1$

$A$ plane ( $\pi$ ) passing through the point $(1, 2, -3)$ is perpendicular to the planes $x + y - z + 4 = 0$ and $2x - y + z + 1 = 0$. If the equation of the plane ( $\pi$ ) is $ax + by + cz + 1 = 0$, then $a^2 + b^2 + c^2 =$

Find the equation of a plane that is at a unit distance from the origin and is parallel to the plane $x - 2y + 2z - 5 = 0$.

$A$ tetrahedron has vertices at $P(2,1,3)$,$Q(-1,1,2)$,$R(1,2,1)$ and $O(0,0,0)$. The angle between the faces $OPQ$ and $PQR$ is

The equation of the plane passing through the point $(2, -1, -3)$ and parallel to the lines $\frac{x - 1}{3} = \frac{y + 2}{2} = \frac{z}{-4}$ and $\frac{x}{2} = \frac{y - 1}{-3} = \frac{z - 2}{2}$ is