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 $\alpha=$

If the equation of the plane which is at a distance of $\frac{1}{3}$ units from the origin and perpendicular to a line whose directional ratios are $(1, 2, 2)$ is $x+py+qz+r=0$,then $\sqrt{p^2+q^2+r^2}=$

The plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$ cuts the $X$-axis at $A$,$Y$-axis at $B$,and $Z$-axis at $C$. The area of $\triangle ABC$ 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:

If the plane $56x + 4y + 9z = 2016$ meets the coordinate axes at points $A$,$B$,and $C$,then the centroid of the $\triangle ABC$ is

If the angle between the planes $x-2y+3z-5=0$ and $x+\alpha y+2z+7=0$ is $\cos^{-1}\left(\frac{1}{14}\right)$,then the difference between the values of $\alpha$ is