In a triangle $ABC$, if the midpoints of sides $AB, BC, CA$ are $(3,0,0), (0,4,0), (0,0,5)$ respectively, then $AB^2+BC^2+CA^2=$

The centroid of the triangle whose vertices are $P(1, -2, 1)$,$Q(2, 3, -1)$,and $R(1, -1, -1)$ is:

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

$ABCD$ 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

In $\triangle ABC$,the midpoints of the sides $AB, BC$ and $CA$ are respectively $(l, 0, 0), (0, m, 0)$ and $(0, 0, n)$. Then,$\frac{AB^2+BC^2+CA^2}{l^2+m^2+n^2}$ is equal to

The points $A(-1, 2, 3)$,$B(2, -3, 1)$,and $C(3, 1, -2)$: