$A$ 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 $\cos \theta =$

If $\alpha$ is the acute angle between the planes $P_1$ and $P_2$,where the combined equation of the planes $P_1$ and $P_2$ is $2x^2 - 6y^2 - 12z^2 + 18yz + 2zx + xy = 0$,then the value of $\cos \alpha$ is:

The direction cosines of the normal to the plane $x + 2y - 3z + 4 = 0$ are

The equation of a plane parallel to the plane $x - 2y + 2z - 5 = 0$ and at a unit distance from the origin is:

Find the equation of the plane bisecting the angle between the planes $2x - y + 2z + 3 = 0$ and $3x - 2y + 6z + 8 = 0$.

Let the plane $\pi$ pass through the point $(1,0,1)$ and be perpendicular to the planes $2x+3y-z=2$ and $x-y+2z=1$. Let the equation of the plane passing through the point $(11,7,5)$ and parallel to the plane $\pi$ be $ax+by-z-d=0$. Then,$\frac{a}{b}+\frac{b}{d}=$