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(A) The centroid $G$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \

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an obtuse angled triangle

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(A) The centroid $G$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.Given $G = \left(5, -1, \frac{2}{3}\right)$,we have:$\frac{1+h-4}{3} = 5$ $\Rightarrow h-3 = 15$ $\Rightarrow h = 18$.$\frac{2-3+k}{3} = -1$ $\Rightarrow k-1 = -3$ $\Rightarrow k = -2$.Thus,the vertices are $A(1, 2, 3)$,$B(18, -3, 0)$,and $C(-4, -2, -1)$.Calculate the squares of the side lengths:$AB^2 = (18-1)^2 + (-3-2)^2 + (0-3)^2 = 17^2 + (-5)^2 + (-3)^2 = 289 + 25 + 9 = 323$.$BC^2 = (-4-18)^2 + (-2-(-3))^2 + (-1-0)^2 = (-22)^2 + 1^2 + (-1)^2 = 484 + 1 + 1 = 486$.$CA^2 = (-4-1)^2 + (-2-2)^2 + (-1-3)^2 = (-5)^2 + (-4)^2 + (-4)^2 = 25 + 16 + 16 = 57$.Since $BC^2 = 486$ and $AB^2 + CA^2 = 323 + 57 = 380$,we observe that $BC^2 > AB^2 + CA^2$.Therefore,the triangle is an obtuse angled triangle.

If the vertices of a triangle $ABC$ are $A(1, 2, 3)$,$B(h, -3, 0)$,and $C(-4, k, -1)$ and the centroid of the triangle is $\left(5, -1, \frac{2}{3}\right)$,then triangle $ABC$ is

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