Let $A(4, -2)$,$B(1, 1)$,and $C(9, -3)$ be the vertices of a triangle $ABC$. Then the maximum area of the parallelogram $AFDE$,formed with vertices $D, E$,and $F$ on the sides $BC, CA$,and $AB$ of the triangle $ABC$ respectively,is $\qquad$ .

Let $A (2, 3, 5)$,$B (-1, 3, 2)$ and $C (\lambda, 5, \mu)$ be the vertices of a $\Delta ABC$. If the median through $A$ is equally inclined to the coordinate axes,then

$A(2, 3, k)$, $B(-1, k, -1)$, and $C(4, -3, 2)$ are the vertices of $\triangle ABC$. If $AB = AC$ and $k > 0$, then $\triangle ABC$ is:

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 $\frac{9 \sqrt{2}}{2}$.

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

Three vertices of a parallelogram $ABCD$ are $A(3, -1, 2)$,$B(1, 2, -4)$,and $C(-1, 1, 2)$. Find the coordinates of the fourth vertex $D$.