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

$P_1$ and $P_2$ are two distinct and intersecting planes. Three non-collinear points lie on $P_1$ and another three non-collinear points lie on $P_2$ (none being on the line of intersection of the planes). Then the maximum number of tetrahedrons formed using these six points is:

$A$ tetrahedron has vertices $O(0,0,0)$, $A(1,2,1)$, $B(2,1,3)$, and $C(-1,1,2)$. If $\theta$ is the angle between the faces $OAB$ and $ABC$, then $\cos \theta =$

The equation of the plane passing through the origin and perpendicular to the planes $x+2y-z=1$ and $3x-4y+z=5$ is:

$A$ vector $\vec{n}$ is inclined to the $x$-axis at $45^\circ$,to the $y$-axis at $60^\circ$,and at an acute angle to the $z$-axis. If $\vec{n}$ is a normal to a plane passing through the point $(\sqrt{2}, -1, 1)$,then the equation of the plane is:

The distance of the plane $\vec{r} = (\hat{i} - \hat{j}) + \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - 2\hat{j} + 3\hat{k})$ from the origin is