The equations of planes parallel to the plane $x+2y+2z+8=0$,which are at a distance of $2$ units from the point $(1,1,2)$ are

$A$ vector $\vec{n}$ is inclined to $X$-axis at $45^{\circ}$,$Y$-axis at $60^{\circ}$ and at an acute angle to $Z$-axis. If $\vec{n}$ is normal to a plane passing through the point $(-\sqrt{2}, 1, 1)$,then the equation of the plane is

$A$ plane meets the coordinate axes at $A, B, C$ respectively such that the centroid of the $\triangle ABC$ is $(2, 3, 5)$. Then, the equation of that plane is

The plane which bisects the line segment joining the points $A(4, -2, 3)$ and $B(2, 4, -1)$ at right angles also passes through the point:

$\vec{n}$ is a unit vector normal to the plane $\pi$ containing the vectors $\hat{i}+3 \hat{k}$ and $2 \hat{i}+\hat{j}-\hat{k}$. If this plane $\pi$ passes through the point $(-3,7,1)$ and $p$ is the perpendicular distance from the origin to this plane $\pi$, then $\sqrt{p^2+5}=$

The length of the perpendicular from the origin to the plane $\bar{r} \cdot (3 \hat{i} - 4 \hat{j} + 12 \hat{k}) = 8$ is