Find the distance between the point $P(6,5,9)$ and the plane determined by points $A(3,-1,2)$,$B(5,2,4)$,and $C(-1,-1,6)$.

Let $P$ be a plane passing through the points $(2,1,0)$,$(4,1,1)$,and $(5,0,1)$,and $R$ be the point $(2,1,6)$. Then the image of $R$ in the plane $P$ is:

The equation of a plane passing through $(-1, 2, 3)$ and whose normal makes equal angles with the coordinate axes is

Let three planes be:
$P_1 : x - y + z = 1$
$P_2 : x + y - z = -1$
$P_3 : x - 3y + 3z = 2$
Let $L_1, L_2, L_3$ be the lines of intersection of planes $(P_2, P_3)$,$(P_3, P_1)$,and $(P_1, P_2)$ respectively.
Statement-$1$: At least two of the lines $L_1, L_2, L_3$ are not parallel.
Statement-$2$: The three planes do not have a common point.

Find the angle between the planes $2x - y + z = 6$ and $x + y + 2z = 3$.

The volume (in cubic units) of the tetrahedron bounded by the plane $3x + 4y - 5z = 60$ and the three coordinate planes is