If $P$ is the point $(2, 6, 3)$,then the equation of the plane passing through $P$ and perpendicular to $OP$,where $O$ is the origin,is:

Let the plane passing through the point $(-1, 0, -2)$ and perpendicular to each of the planes $2x + y - z = 2$ and $x - y - z = 3$ be $ax + by + cz + 8 = 0$. Then the value of $a + b + c$ is equal to:

Let $S$ be the set of all real values of $\lambda$ such that a plane passing through the points $(-\lambda^2, 1, 1)$,$(1, -\lambda^2, 1)$,and $(1, 1, -\lambda^2)$ also passes through the point $(-1, -1, 1)$. Then $S$ is equal to

$A$ 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 the foot of the perpendicular drawn from the point $(1,0,-2)$ to the plane $\pi$ is $(2,0,-1)$ and the equation of the plane $\pi$ is $ax+by+cz=2$, then $a^2+b^2+c^2=$

The direction ratios of the lines $OA$ and $OB$ are $1, -2, -1$ and $3, -2, 3$. Then the direction cosines of the normal of plane $AOB$,where $O$ is the origin,are