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

If $P(0, 7, 10)$,$Q(-1, 6, 6)$,and $R(-4, 9, 6)$ are three points in space,then $\triangle PQR$ is:

If the points $P = \hat{i} + 2 \hat{j}$,$Q = 4 \hat{i} + 6 \hat{j}$,$R = 5 \hat{i} + 7 \hat{j}$,and $S = a \hat{i} + b \hat{j}$ are the consecutive vertices of a parallelogram $PQRS$,then:

The points $A(2, 3, 5)$,$B(-1, 5, -1)$,and $C(4, -3, 2)$ form:

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

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