Find the equation of a plane which bisects the line segment joining the points $A(2,3,4)$ and $B(4,5,8)$ at right angles.

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 $p - q =$

The direction cosines of the normal to the plane $x + 2y - 3z + 4 = 0$ are

Find the equation of the plane that passes through the three points $(1, 1, -1)$,$(6, 4, -5)$,and $(-4, -2, 3)$.

If $A$ and $B$ are the feet of the perpendiculars drawn from point $Q(a, b, c)$ to the planes $yz$ and $zx$ respectively,then the equation of the plane passing through the points $A, B$ and the origin $O(0, 0, 0)$ is $.........$

The distance of the point $(2, -1, 0)$ from the plane $2x + y + 2z + 8 = 0$ is