The Cartesian equation of the plane,passing through the points $(3,1,1)$,$(1,2,3)$ and $(-1,4,2)$,is

$A$ plane containing two lines whose direction ratios are $(-1, 2, 1)$ and $(1, 3, 2)$ passes through the point $(2, 1, k)$. If this plane also passes through the point $(3, -1, 4)$,then $k=$

$A$ 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):

$(1, -2, 1)$ is a point on a plane $\pi$ and $\pi$ is parallel to the plane $x-y-z=0$. If the equation of $\pi$ is $ax+by+cz-2=0$,then $b-2c=$

Let the image of the point $P(1, 2, 6)$ in the plane passing through the points $A(1, 2, 0)$,$B(1, 4, 1)$,and $C(0, 5, 1)$ be $Q(\alpha, \beta, \gamma)$. Then $(\alpha^2 + \beta^2 + \gamma^2)$ is equal to :

$A$ plane passing through the points $(0, -1, 0)$ and $(0, 0, 1)$ and making an angle $\frac{\pi}{4}$ with the plane $y - z + 5 = 0$ also passes through the point