The perpendicular distance of the origin from the plane $x-3y+4z-6=0$ 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 vector equation of the plane which is at a distance of $ \frac{3}{\sqrt{14}} $ from the origin and the normal vector from the origin is $ 2 \hat{i}-3 \hat{j}+\hat{k} $ is:

The equation of the plane passing through the point $(1, 2, -3)$ and perpendicular to the planes $3x + y - 2z = 5$ and $2x - 5y - z = 7$ is:

Find the equation of a plane that is at a unit distance from the origin and is parallel to the plane $x - 2y + 2z - 5 = 0$.

$\vec{n}$ is a unit vector normal to the plane $\pi$ containing the vectors $\hat{i}+3 \hat{k}$ and $2 \hat{i}+\hat{j}-\hat{k}$. If this plane $\pi$ passes through the point $(-3,7,1)$ and $p$ is the perpendicular distance from the origin to this plane $\pi$,then $\sqrt{p^2+5}=$