The points $(5, -4, 2), (4, -3, 1), (7, -6, 4)$ and $(8, -7, 5)$ are the vertices of

If $P(3,2,6), Q(1,4,5)$ and $R(3,5,3)$ are the vertices of $\Delta PQR$,then $m \angle PQR$ is (in $^{\circ}$)

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

If the centroid of a triangle with vertices $(4, p, -3)$,$(-1, -1, 2)$,and $(3, 5, -8)$ is given by the mid-point of $(1, 4, -2)$ and $(q, 2, -4)$,then $p^2 + q^2 =$

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

If $A(1,4,2)$ and $C(5,-7,1)$ are two vertices of triangle $ABC$ and $G\left(\frac{4}{3}, 0, \frac{-2}{3}\right)$ is the centroid of the triangle $ABC$,then the midpoint of side $BC$ is