The mirror image of the point $P(-1, 2, -4)$ in the plane $x - y - 2z + 1 = 0$ is

If $(2, -1, 3)$ is the foot of the perpendicular drawn from the origin $(0, 0, 0)$ to a plane,then the equation of that plane is:

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 $p - q =$

$A$ tetrahedron has vertices $O(0,0,0)$, $A(1,2,1)$, $B(2,1,3)$, and $C(-1,1,2)$. If $\theta$ is the angle between the faces $OAB$ and $ABC$, then $\cos \theta =$

$A$ plane passes through the point $A(2, 1, -3)$. If the distance of this plane from the origin is maximum,then its equation is

If $P$ is the point $(2, 6, 3)$,then the equation of the plane passing through $P$ and perpendicular to $OP$,where $O$ is the origin,is: