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

The points $(-2, 4, 7)$,$(3, -6, -8)$,and $(1, -2, -2)$ are:

The points $(5,-4,5), (-3,-3,2)$ and $(-1,-6,8)$ form ......

Points $(1, 1, 1), (-2, 4, 1), (-1, 5, 5)$ and $(2, 2, 5)$ are the vertices of a

The points $(2, 3, 4)$,$(-1, -2, 1)$,and $(5, 8, 7)$ are

Let $A(x, y, z)$ be a point in the $xy$-plane,which is equidistant from three points $P(0, 3, 2)$,$Q(2, 0, 3)$,and $R(0, 0, 1)$. Let $B = (1, 4, -1)$ and $C = (2, 0, -2)$. Then among the statements $(S1) :$ $\triangle ABC$ is an isosceles right-angled triangle and $(S2) :$ the area of $\triangle ABC$ is $\frac{9 \sqrt{2}}{2}$.