If the orthocentre and the centroid of a triangle are at $(5,2,-6)$ and $(9,6,-4)$ respectively,then its circumcentre is

  • A
    $(11,8,-3)$
  • B
    $(8,8,-3)$
  • C
    $(11,8,3)$
  • D
    $(11,-8,-3)$

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Find the distance of the point $P(3, 8, 2)$ from the line $\frac{x-1}{2} = \frac{y-3}{4} = \frac{z-2}{3}$ measured parallel to the plane $3x + 2y - 2z + 15 = 0$.

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$A$ straight line drawn from the point $P(1,3,2)$,parallel to the line $\frac{x-2}{1}=\frac{y-4}{2}=\frac{z-6}{1}$,intersects the plane $L_1: x-y+3z=6$ at the point $Q$. Another straight line which passes through $Q$ and is perpendicular to the plane $L_1$ intersects the plane $L_2: 2x-y+z=-4$ at the point $R$. Then which of the following statements is(are) $TRUE$?
$(A)$ The length of the line segment $PQ$ is $\sqrt{6}$
$(B)$ The coordinates of $R$ are $(1,6,0)$
$(C)$ The centroid of the triangle $PQR$ is $\left(\frac{4}{3}, \frac{14}{3}, \frac{5}{3}\right)$
$(D)$ The perimeter of the triangle $PQR$ is $\sqrt{6}+\sqrt{13}+\sqrt{11}$

The incentre of the triangle whose vertices are $P(0,3,0)$,$Q(0,0,4)$,and $R(0,3,4)$ is

What is the angle subtended by an edge of a regular tetrahedron at its centre?

Let $A(1, 2, 3)$, $B(-1, 4, 6)$, $C(0, -6, 4)$, and $D(1, 1, 1)$ be the vertices of a tetrahedron. Let $G$ be its centroid and $G_1$ be the centroid of its face $BCD$. Then $\frac{AG_1}{AG} =$

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