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

$A$ vector $n$ of magnitude $8$ units is inclined to $x$-axis at $45^\circ$,$y$-axis at $60^\circ$ and an acute angle with $z$-axis. If a plane passes through a point $(\sqrt{2}, -1, 1)$ and is normal to $n$,then its equation in vector form is

The distance of the plane $6x - 3y + 2z - 14 = 0$ from the origin is

Let $P$ be the plane passing through the point $(1, 2, 3)$ and the line of intersection of the planes $\vec{r} \cdot (\hat{i} + \hat{j} + 4\hat{k}) = 16$ and $\vec{r} \cdot (-\hat{i} + \hat{j} + \hat{k}) = 6$. Then which of the following points does $NOT$ lie on $P$?

Find the equation of the planes bisecting the angles between the planes $\vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) = 19$ and $\vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\hat{k}) + 3 = 0$.

Find the angle between the planes $x+2y+2z-5=0$ and $3x+3y+2z-8=0$.