The equation of the plane passing through the points $(2, 3, 1)$ and $(4, -5, 3)$ and parallel to the $X$-axis is:

In three-dimensional space,if $f(x, y, z) = xy + xz$,then the locus of all points which satisfy the equation $f(x, y, z) = 0$ is -

If the plane $\frac{x}{3}+\frac{y}{2}-\frac{z}{4}=1$ cuts the coordinate axes at points $A, B$,and $C$,then the area of the triangle $ABC$ is

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:

Find the equation of a plane,given that the foot of the perpendicular drawn to the plane from the origin is $(2, 1, 2)$.

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 $B(1,3,4).$