The equation of the plane which passes through $(2, -3, 1)$ and is normal to the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$ is given by

The equation of the plane passing through the point $(2, -1, -3)$ and parallel to the lines $\frac{x - 1}{3} = \frac{y + 2}{2} = \frac{z}{-4}$ and $\frac{x}{2} = \frac{y - 1}{-3} = \frac{z - 2}{2}$ is

$A$ plane is parallel to two lines,whose direction ratios are $1, 0, -1$ and $-1, 1, 0$ and it contains the point $(1, 1, 1)$. If it cuts coordinate axes ($X, Y, Z$-axes respectively) at $A, B, C$,then the volume of the tetrahedron $OABC$ is (in cubic units):

Let $P(1, -2, 5)$ be the foot of the perpendicular drawn from the origin to the plane $\pi_1$ and the same $P$ be the foot of the perpendicular from $(1, 2, -1)$ to the plane $\pi_2$. Then the acute angle between the planes $\pi_1$ and $\pi_2$ 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:

Let $\bar{n}$ be a vector of magnitude $3\sqrt{3}$ such that it makes equal acute angles with the coordinate axes. Then the vector equation of a plane passing through $(1, -1, 2)$ and normal to $\bar{n}$ is: