$A$ plane $\pi$ passes through the points $(5,1,2)$,$(3,-4,6)$,and $(7,0,-1)$. If $p$ is the perpendicular distance from the origin to the plane $\pi$ and $l, m, n$ are the direction cosines of a normal to the plane $\pi$,then $|3l+2m+5n|=$

The vector equation of a plane,which is at a distance of $8$ units from the origin and which is normal to the vector $\vec{n} = 2\hat{i} + \hat{j} + 2\hat{k}$,is

Equations of planes parallel to the plane $x-2y+2z+4=0$ which are at a distance of $1$ unit from the point $(1, 2, 3)$ are $.....$

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 $ax + by \pm (\sqrt{a^{2} + b^{2}} \tan \alpha) z = 0$.

$A$ plane $\pi_1$ passing through the point $3 \hat{i}-7 \hat{j}+5 \hat{k}$ is perpendicular to the vector $\hat{i}+2 \hat{j}-2 \hat{k}$ and another plane $\pi_2$ passing through the point $2 \hat{i}+7 \hat{j}-8 \hat{k}$ is perpendicular to the vector $3 \hat{i}+2 \hat{j}+6 \hat{k}$. If $p_1$ and $p_2$ are the perpendicular distances from the origin to the planes $\pi_1$ and $\pi_2$ respectively,then $p_1-p_2=$

The distance between the planes given by $r \cdot (i + 2j - 2k) + 5 = 0$ and $r \cdot (i + 2j - 2k) - 8 = 0$ is