Find the equation of the plane passing through the point $(2, 3, 1)$ and perpendicular to the line $\frac{x - 1}{1} = \frac{y - 2}{-1} = \frac{z + 1}{2}$.

$A$ plane $P$ is parallel to two lines whose direction ratios are $-2, 1, -3$ and $-1, 2, -2$,and it contains the point $(2, 2, -2)$. Let $P$ intersect the coordinate axes at the points $A, B, C$ making the intercepts $\alpha, \beta, \gamma$. If $V$ is the volume of the tetrahedron $OABC$,where $O$ is the origin and $p = \alpha + \beta + \gamma$,then the ordered pair $(V, p)$ is equal to.

The equation of a 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 direction ratios of the normal to the plane passing through $(0,0,1)$,$(0,1,2)$,and $(1,0,3)$ are:

If the image of the point $A(1, 1, 1)$ with respect to the plane $4x + 2y + 4z + 1 = 0$ is $B(\alpha, \beta, \gamma)$,then $\alpha + \beta + \gamma =$

The image point of $(1, 3, 4)$ in the plane $2x - y + z + 3 = 0$ is