The equation of the plane which bisects the line joining $(2, 3, 4)$ and $(6, 7, 8)$ is

The equation of the plane passing through $(-1, 1, 2)$ whose normal makes equal acute angles with the coordinate axes is:

The plane $ax + by = 0$ is rotated about its line of intersection with the plane $z = 0$ through an angle $\alpha$. Prove that the equation of the plane in its new position is $ax + by \pm (\sqrt{a^{2} + b^{2}} \tan \alpha) z = 0$.

The equation of the plane passing through the point $(1,2,2)$ and perpendicular to the planes $x-y+2z=3$ and $2x-2y+z+12=0$ is:

If the angle between the planes $\bar{r} \cdot(11 \hat{i}-2 \hat{j}+\alpha \hat{k})=7$ and $\bar{r} \cdot(2 \hat{i}+4 \hat{j}-2 \hat{k})=5$ is $\frac{\pi}{2}$, then $\alpha=$

The Cartesian equation of the plane passing through the point $A(7,8,6)$ and parallel to the $XY$ plane is