The triangle formed by the points $(0, 7, 10), (-1, 6, 6), (-4, 9, 6)$ is

Find the third vertex of a triangle whose centroid is the origin and two vertices are $(2, 4, 6)$ and $(0, -2, -5)$.

$ABCD$ is a parallelogram,$P$ is the mid-point of $AB$. If $R$ is the point of intersection of $AC$ and $DP$,then $R$ divides $AC$ internally in the ratio

If $a>0, b>0$, then the maximum area of the parallelogram whose three vertices are $O(0,0)$, $A(a \cos \theta, b \sin \theta)$, and $B(a \cos \theta, -b \sin \theta)$ is

If the three vertices of a parallelogram are $(-1, 0), (3, 1),$ and $(2, 2)$ respectively,find the coordinates of the fourth vertex.

If $P(3,2,6), Q(1,4,5)$ and $R(3,5,3)$ are the vertices of $\Delta PQR$,then $m \angle PQR$ is (in $^{\circ}$)