The circumradius of the triangle formed by the points $A(2, -1, 1)$,$B(1, -3, -5)$,and $C(3, -4, -4)$ 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

The perimeter of the triangle with vertices at $(1,0,0), (0,1,0)$ and $(0,0,1)$ is

$\triangle ABC$ has vertices at $A \equiv (2, 3, 5)$,$B \equiv (-1, 3, 2)$ and $C \equiv (\lambda, 5, \mu)$. If the median through $A$ is equally inclined to the axes,then the values of $\lambda$ and $\mu$ respectively are

Verify that the points $(0, 7, 10)$,$(-1, 6, 6)$,and $(-4, 9, 6)$ are the vertices of a right-angled triangle.

Show that the points $A(1, 2, 3)$,$B(-1, -2, -1)$,$C(2, 3, 2)$,and $D(4, 7, 6)$ are the vertices of a parallelogram $ABCD$,but it is not a rectangle.