Direction ratios of the normal to the plane passing through the points $A(0, 1, 1)$,$B(1, 1, 2)$,and $C(-1, 2, -2)$ are

Equation of planes parallel to the plane $x-2y+2z+4=0$ which are at a distance of one unit from the point $(1,2,3)$ are

From a point $P(a, b, c)$, perpendiculars $PA$ and $PB$ are drawn to $XY$ plane and $ZX$ plane respectively. If $O$ is the origin, then the equation of plane $OAB$ is

The foot of the perpendicular drawn from the point $(-2,-1,3)$ to a plane $\pi$ is $(1,0,-2)$. If $a, b, c$ are the intercepts made by the plane $\pi$ on $X, Y, Z$-axes respectively, then $3a+b+5c=$

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 $p - q =$

Let $P$ be the plane passing through the point $(1, -1, -5)$ and perpendicular to the line joining the points $(4, 1, -3)$ and $(2, 4, 3)$. Then the distance of $P$ from the point $(3, -2, 2)$ is