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  4. A= (1, -2, 3), B = (2, 1, 3), C = (4, 2, 1) and G = (-1, 3, 5) is the centroid of the tetrahedron ABCD. If p= Dy and q = Dz then 13p-1.1q =

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Q. $A= (1, -2, 3), B = (2, 1, 3), C = (4, 2, 1)$ and $G = (-1, 3, 5)$ is the centroid of the tetrahedron ABCD. If $p= D_y$ and $q = D_z$ then $ 13p-1.1q =$

Three Dimensional Geometry

Solution:

$\left(\frac{1+2+4+x}{4},\frac{-2 +1 +2 +y }{4},\frac{3+3+1+z}{4}\right)$
$= (-1 ,3,5)$