The vector equation of the plane passing through the origin and the line of intersection of the planes $r \cdot a = \lambda$ and $r \cdot b = \mu$ is

If $a, b, c$ are the intercepts made on $X, Y, Z$-axes respectively by the plane passing through the points $(1, 0, -2), (3, -1, 2)$ and $(0, -3, 4)$, then $3a + 4b + 7c =$

Statement $-1:$ The point $A(3,1,6)$ is the mirror image of the point $B(1,3,4)$ in the plane $x-y+z=5.$
Statement $-2:$ The plane $x-y+z=5$ bisects the line segment joining $A(3,1,6)$ and $B(1,3,4).$

If two planes intersect,then the shortest distance between the planes is

Let $\pi_1$ be the plane passing through the points $(0,1,2), (1,0,-2), (-2,1,0)$ and $\pi_2$ be the plane passing through the point $(1,2,3)$ and perpendicular to the planes $x+y+z=1$ and $2x-3y+z=5$. If $\theta$ is the acute angle between the planes $\pi_1$ and $\pi_2$, then $\cos \theta=$

$A$ plane meets the coordinate axes at points $A$, $B$ and $C$, such that the centroid of triangle $ABC$ is $(2, -\frac{2}{3}, \frac{1}{2})$. The perpendicular distance from the origin to this plane is: