If the mid-points of the sides $AB, BC, CA$ of a triangle are $(1, 5, -1), (0, 4, -2), (2, 3, 4)$ respectively,then the length of the median drawn from $C$ to $AB$ is

The perimeter of the triangle with vertices at $(1,0,0), (0,1,0)$ and $(0,0,1)$ 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 $\frac{9 \sqrt{2}}{2}$.

$A$ point moves in such a way that the sum of its distances from the $xy$-plane and $yz$-plane remains equal to its distance from the $zx$-plane. The locus of the point is:

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

Given $\triangle ABC$ such that $A = 2\hat{i} - \hat{j} + \hat{k}$,$B = \hat{i} - 3\hat{j} - 5\hat{k}$,and $C = 3\hat{i} - 4\hat{j} - 4\hat{k}$,then $\triangle ABC$ is: